PRACTICE PROBLEMS: SET 6
MATH 101: PROF. DRAGOS GHIOCA
Problem 1. Compute
∫
1. Problems
sec(x) dx.
Problem 2. Compute
∫
tan2 (x) sec(x) dx .
Problem 3. Evaluate
∫0
−1
1
(x2 + 4x + 3) 2 dx.
Problem 4. Compute
∫
x3
Problem 5. Evaluate
∫
x2 + 2
dx .
+ 6x + 1
x2 + x + 1
dx .
x3 − 2x2 − x + 2
Problem 6. Compute
∫
Problem 7. Compute
∫
Problem 8. Compute
∫
x+2
dx .
x3 + x2
x2 − 1
dx .
x3 + 3x2 + 5x
2x2 − 4x + 5
dx .
x4 + x3 + 2x2
Problem 9. Compute
∫
sec3 (x) dx .
Problem 10. Compute
∫
Problem 11. Compute
Problem 12. Compute
∫
∫
√
3
x
dx .
x−1
√
sin( x) dx .
x2 + 1
dx .
x2 + x + 1
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1. This is not a rational function. However, its integral is done in two
steps. The second step will involve partial fractions. ∫
∫
The first step is an algebraic manipulation. We have sec(x) dx = (cos(x))−1 dx.
Therefore, we integrate an odd power of the cosine (−1 is an odd integer; it does not
matter that is negative). We will integrate with the usual method for trigonometric
integrals involving an odd power, i.e. we isolate one positive power of the cosine.
Hence
∫
∫
∫
cos(x) dx
cos(x)−1 dx = cos(x)−2 · cos(x) dx =
.
cos2 (x)
Then we will attempt a substitution for which du = cos(x) dx. This means u =
sin(x) and so, we need to express cos2 (x) in terms of sin(x). Thus
∫
cos(x) dx
cos2 (x)
∫
cos(x) dx
=
1 − sin2 (x)
∫
du
=
.
1 − u2
Now we can integrate using partial fractions. We have
∫
∫
du
du
=
−
.
1 − u2
u2 − 1
A
B
+ u+1
. We compute A and B
For the fraction u21−1 the partial fractions are u−1
as in class and get (after bringing everything to a common denominator):
1 = A(u + 1) + B(u − 1).
We plug in u = 1 in the above equality and get 1 = A · 2 + 0 and so, A = 21 .
Similarly, we plug in u = −1 and get 1 = 0 + B · (−2) and so, B = −1
2 . Therefore
∫
∫
1
1
du
1
1
2
2
=
−
du = ln | u − 1 | − | u + 1 | + C .
2
u −1
u−1 u+1
2
2
We plug in back u = sin(x) and conclude
∫
sec(x) dx
∫
du
= −
u2 − 1
1
1
= − ln | u − 1 | + ln | u + 1 | + C
2
2
1
1
= − ln | sin(x) − 1 | + ln | sin(x) + 1 | + C
2 2
1 sin(x) + 1 =
ln
+C.
2 sin(x) − 1 We use algebraic manipulations in the last formula to express it slightly different.
sin(x)+1
with sin(x) + 1. We
So, we multiply both numerator and denominator of sin(x)−1
PRACTICE PROBLEMS: SET 6
get
2
2
(sin(x) + 1)
(sin(x) + 1)
=
=−
2
− cos2 (x)
sin (x) − 1
Thus
sin(x) + 1 ln sin(x) − 1 =
=
=
=
So, we conclude
(
sin(x) + 1
cos(x)
3
)2
.
(
)2 sin(x) + 1 ln −
cos(x)
2
sin(x) + 1 ln cos(x) sin(x)
1 2 ln +
cos(x) cos(x) 2 ln | tan(x) + sec(x) | .
∫
sec(x) dx
1 sin(x) + 1 ln +C
2
sin(x) − 1 = ln | tan(x) + sec(x) | + C .
=
Problem 2. For this integral we need to be inventive since it does not contain
an even power of sec(x), nor it contains an odd power of tan(x). So, we use first
the trigonometric identity:
tan2 (x) = sec2 (x) − 1 and so,
∫
∫
∫
∫
tan2 (x) sec(x) dx = (sec2 (x) − 1) · sec(x) dx = sec3 (x) dx − sec(x) dx .
We already computed the antiderivative of sec(x) in Problem 1 and obtained:
∫
(1)
sec(x) dx = ln | tan(x) + sec(x) | + C .
∫
So, now we have to evaluate sec3 (x) dx. Here we use integration by parts:
∫
sec3 (x) dx
∫
=
sec(x) · sec2 (x) dx
∫
=
sec(x) · d (tan(x))
∫
= sec(x) · tan(x) − tan(x)d (sec(x))
∫
= tan(x) sec(x) − tan(x) · tan(x) sec(x) dx
∫
= tan(x) sec(x) − tan2 (x) sec(x) dx .
4
MATH 101: PROF. DRAGOS GHIOCA
Therefore
∫
∫
sec3 (x) dx = tan(x) sec(x) −
(2)
tan2 (x) sec(x) dx .
Now, we recall from where we started:
∫
∫
∫
tan2 (x) sec(x) dx = sec3 (x) dx − sec(x) dx .
So, substituting in (2) the relation:
∫
∫
∫
sec3 (x) dx = tan2 (x) sec(x) dx + sec(x) dx,
we obtain
∫
∫
2
tan (x) sec(x) dx + sec(x) dx
∫
= tan(x) sec(x) −
tan2 (x) sec(x) dx,
∫
and so, solving for tan2 (x) sec(x) dx we obtain that
∫
∫
2 tan2 (x) sec(x) dx = tan(x) sec(x) − sec(x) dx .
Therefore
∫
tan2 (x) sec(x) dx
∫
tan(x) sec(x) 1
− · sec(x) dx
=
2
2
tan(x) sec(x) − ln | tan(x) + sec(x) |
=
+ C,
2
where for the last equality we also used (1).
1
Problem 3. We will compute the antiderivative of (x2 + 4x + 3) 2 and only at the
end we will evaluate the definite integral.
We first complete the square in order to bring the expression under the radical
to one of the three forms ±a2 ± x2 . So,
x2 + 4x + 3 = (x + 2)2 − 1.
Then we make the change of variable u = x + 2, which means du = dx. So, we
obtain
∫
∫
1
1
2
2
(x + 4x + 3) dx = (u2 − 1) 2 du .
This last integral involves a trigonometric substitution of u = sec(t) and thus
du = tan(t) sec(t) dt. Using that u2 − 1 = sec2 (t) − 1 = tan2 (t), we get
∫ √
u2 − 1 du
∫
=
tan(t) · tan(t) sec(t) dt
∫
=
tan2 (t) sec(t) dt .
PRACTICE PROBLEMS: SET 6
5
But we alreay computed the last integral in Problem 2 and we obtained
∫
tan(t) sec(t) ln | sec(t) + tan(t) |
tan2 (t) sec(t) dt =
−
+ C,
2
2
which leads us to conclude that
∫
1
1
1
u(u2 − 1) 2
ln | u + (u2 − 1) 2 |
(u2 − 1) 2 du =
−
+C.
2
2
√
(For the above formula note that sec(t) = u and also, tan(t) = u2 − 1.)
Changing back to x (and using that u = x + 2), we get that our definite integral
equals
(
)
1
1
ln | x + 2 + ((x + 2)2 − 1) 2 | 0
(x + 2)((x + 2)2 − 1) 2
−
|−1
2
2
√
√
2 3 − ln(2 + 3)
=
.
2
Problem 4. The numerator has degree less than the denominator; so, there is no
point to do any long division (we would get the same fraction through long division).
We can easily integrate the fraction if we observe that actually, the numerator and
the denominator are in a very close relation.
We observe that if we attempt a substitution: u = x3 + 6x + 1, then
du = (3x2 + 6) dx = 3(x2 + 2) dx .
Hence,
∫
x3
x2 + 2
dx
+ 6x + 1
∫
=
du
3
u
ln | u |
=
+C
3
ln | x3 + 6x + 1 |
=
+C.
3
Even though it happens very rarely that we are so lucky to have such
a relation between numerator and denominator, i.e. the numerator is a
constant multiple of the derivative of the denominator, we should first
check that indeed this is the case before we attempt something more
elaborate as partial fractions.
Problem 5. We observe that
x3 − 2x2 − x + 2 = (x − 2)(x2 − 1) = (x − 2)(x − 1)(x + 1).
The theory says that we can write
x2 +x+1
x3 −2x2 −x+2
as
B
C
A
+
+
.
x−1 x+1 x−2
The whole point of the method is to calculate A, B and C and then to evaluate
the much simpler integrals:
∫
A
dx = A ln | x − 1 | + C
x−1
6
MATH 101: PROF. DRAGOS GHIOCA
∫
B
dx = B ln | x + 1 | + C
x+1
∫
C
dx = C ln | x − 2 | + C .
x−2
We have
x2 + x + 1
A
B
C
=
+
+
.
− 2x2 − x + 2
x−1 x+1 x−2
Hence, after bringing all fractions to the same denominator, we conclude that
x3
x2 + x + 1 = A(x + 1)(x − 2) + B(x − 1)(x − 2) + C(x − 1)(x + 1).
Then, to find out A, we plug in x = 1 in both sides of the above equality and
obtain:
3 = A · 2 · (−1) + 0 + 0.
3
Hence A = − 2 . Similarly, to find out B, we plug in x = −1 and conclude
1 = B · (−2) · (−3).
1
6.
So, B =
Finally, to find out C, we plug in x = 2 and obtain 7 = 3C. Thus
C = 73 . We conclude that
∫
x2 + x + 1
3
1
7
dx = − ln | x − 1 | + ln | x + 1 | + ln | x − 2 | + C .
3
x − 2x2 − x + 2
2
6
3
Problem 6. Clearly, the denominator factors out as x3 + x2 = x2 (x + 1). Therefore, we want to find the corresponding constants for
x+2
A
B
C
= + 2+
.
x3 + x2
x
x
x+1
We bring again all fractions to the same denominator and obtain
x + 2 = A · x(x + 1) + B · (x + 1) + C · x2 .
We let x = −1 in the above equality (in both sides). We obtain C = 1. On the
other hand, if we plug in x = 0 in both sides of the above equality, we obtain B = 2.
Hence,
x + 2 = A(x2 + x) + 2(x + 1) + x2 .
After moving 2(x + 1) + x2 on the left hand side and expanding out and then
collecting the terms, we conclude
−x2 − x = Ax2 + Ax.
This equality of polynomials forces A = −1. (Note that A = −1 is obtained from
the information from both monomials x and x2 in the above equality; this is an
indication that our computation was correct.) So,
∫
x+2
dx
x3 + x2
∫
∫
∫
−1
2
1
=
dx +
dx
+
dx
x
x2
x+1
2
= − ln | x | − + ln | x + 1 | + C .
x
Problem 7. Clearly,
x3 + 3x2 + 5x = x(x2 + 3x + 5) and so,
PRACTICE PROBLEMS: SET 6
7
x2 − 1
A
Bx + C
= + 2
.
+ 3x2 + 5x
x
x + 3x + 5
(Note that x2 +3x+5 is a non-factorable quadratic since its discriminant 32 −4·5 =
−11 is negative.)
We bring all fractions to a common denominator and obtain
x3
x2 − 1 = A · (x2 + 3x + 5) + (Bx + C) · x.
We expand the right hand side and collect terms:
x2 − 1 = (A + B)x2 + (3A + C)x + 5A.
The above equality of polynomials can only occur if the coefficient of each monomial
on the right hand side equals the coefficient of the same monomial on the left hand
side. Hence
x2 − 1 = (A + B)x2 + (3A + C)x + 5A,
and so,
1=A+B
0 = 3A + C
−1 = 5A
−1
5 .
The last equation gives us A =
Using this information in each of the first two
equations, we obtain B = 56 and C = 35 . Therefore
∫
x2 − 1
dx
3
x + 3x2 + 5x
∫
∫
3
6
−1
5x + 5
=
dx +
dx .
2
5x
x + 3x + 5
The first integral clearly equals − 15 ln | x | + C, while the second integral is harder.
One way, but definitely not easy, would be to complete the square in the denominator
(
)2
3
11
x2 + 3x + 5 = x +
+
2
4
and then apply a trigonometric substitution
( ) 12
3
11
tan(t).
x+ =
2
4
This is a good exercise to see how well you work with trigonometric
substitutions!
∫ 6 x+ 35
Another way to compute x25+3x+5
dx would be the following.
First you pretend you want to make a substitution u = x2 + 3x + 5. Hence
du = (2x + 3) dx.
Then compare the numerator of your fraction with what you would like to have,
i.e. we would like very much that the numerator is a constant multiple of (2x + 3).
But it is not that nice!
Still, you can express how far is the numerator from what you would like it to
be.
3
3
6
6
x + = · (2x + 3) − .
5
5
5
5
8
MATH 101: PROF. DRAGOS GHIOCA
Hence
∫
+ 35
dx
+ 3x + 5
6
5x
x2
∫
=
· (2x + 3)
dx −
x2 + 3x + 5
3
5
∫
6
5
x2
+ 3x + 5
dx .
The first integral is solved through a substitution (after all, this was the whole
point of the above splitting of our integral) and we obtain
∫ 3
3
5 · (2x + 3)
dx = ln | x2 + 3x + 5 | + C .
2
x + 3x + 5
5
For the second integral from above, we first complete the square in the denominator and get
∫
1
6
dx .
(
)
3 2
5
x + 2 + 11
4
∫
The above integral looks like u21+1 du = arctan(u) + C. So, we force it to be
exactly that. For this we make the substitution
( ) 12
3
11
x+ =
u.
2
4
Then
(
)2
3
11
11
x+
+
=
· (u2 + 1)
2
4
4
( ) 12
and also, dx = 11
du. So,
4
∫
∫
6
6
6
1
1
5
5
dx
=
du
=
(
)
1
(
)
(
) 1 arctan(u) + C .
2
5
u2 + 1
11 2
11 2
x + 32 + 11
4
4
4
We conclude
∫
6
1
(
)
3 2
5
x+ 2 +
11
4
dx

=

x + 23
( 11 ) 12 · arctan  ( 11 ) 12  + C .
6
5
4
4
Finally, we get that
∫
x2 − 1
dx
3
x + 3x2 + 5x
1
3
12
= − ln | x | + ln | x2 + 3x + 5 | − √ · arctan
5
5
5 11
(
2x + 3
√
11
Problem 8. The denominator factors as x2 (x2 + x + 2), i.e., a repeated linear
factor and a non-factorable quadratic. So, the partial fractions are
2x2 − 4x + 5
A
B
Cx + D
= + 2+ 2
.
x4 + x3 + 2x2
x
x
x +x+2
We obtain
2x2 − 4x + 5 = Ax(x2 + x + 2) + B(x2 + x + 2) + (Cx + D) · x2 .
)
+C.
PRACTICE PROBLEMS: SET 6
9
We plug in x = 0 and we find:
5
.
2
in the above identity of polynomials and obtain
5 = B · 2; therefore B =
We substitute B =
5
2
5
2x2 − 4x + 5 = Ax(x2 + x + 2) + (x2 + x + 2) + (Cx + D) · x2 ,
2
and so,
13
1
− x2 − x = (A + C)x3 + (A + D)x2 + 2Ax.
2
2
Therefore, looking at the coefficient of x in both sides of the above identity, we
obtain
13
A=− .
4
Equating the coefficients of x2 in both sides of the above identity, and also using
that A = − 13
4 , we get
1 13
11
D=− +
=
.
2
4
4
Finally, equating the coefficient of x3 from both sides we conclude that
13
C=
.
4
So,
∫
2x2 − 4x + 5
dx
x4 + x3 + 2x2
∫
∫
∫ 13
11
13
1
5
1
4 x+ 4
= −
dx +
dx
+
dx
4
x
2
x2
x2 + x + 2
∫ 13
11
13
5 1
4 x+ 4
= − ln | x | − · +
dx .
4
2 x
x2 + x + 2
11
We are left to evaluate the last integral. For this we express first 13
4 x + 4 in terms
2
of the derivative of x + x + 2, i.e.,
13
11
13
9
x+
=
· (2x + 1) + .
4
4
8
8
We do this so that we can split the integral
∫ 13
11
4 x+ 4
dx
x2 + x + 2
∫
∫
2x + 1
9
1
13
·
dx + ·
dx .
=
8
x2 + x + 2
8
x2 + x + 2
Then the first integral is immediately computed with the substitution
u = x2 + x + 2 and so, du = (2x + 1) dx .
∫
2x + 1
du
dx =
= ln | u | + C = ln(x2 + x + 2) + C .
x2 + x + 2
u
(Note that in the above expression we used the fact that x2 + x + 2 > 0 always and
thus we don’t need to keep the absolute values.)
Thus
∫
10
MATH 101: PROF. DRAGOS GHIOCA
For the second integral, we complete the square first
)2
(
7
1
2
+ .
x +x+2= x+
2
4
Then we use the substitution
1
x+ =
2
So,
∫
x2
√
√
7
7
u and so, dx =
du .
4
2
1
dx
+x+2
√
∫
=
√
=
=
=
∫
7
2
7
4
7
2 du
7
7
2
4 ·u + 4
∫
·
du
+1
u2
2
√ arctan(u) + C
7
(
)
x + 21
2
√
√ arctan
+C.
7
7
2
We finally conclude that
2x2 − 4x + 5
dx
x4 + x3 + 2x2
13
5
13
9
= − ln | x | −
+
ln(x2 + x + 2) + √ arctan
4
2x
8
4 7
(
2x + 1
√
7
Problem 9. We computed the above integral also in Problem 2 using
the method of trigonometric integrals coupled with integration by parts.
This time we will make a substitution, and then use the method for
rational functions.
We rewrite the integral as
∫
∫
∫
∫
1
cos(x)
cos(x) dx
sec3 (x) dx =
dx
=
dx
=
.
3
2
2
cos (x)
(cos (x))
(1 − sin2 (x))2
We make the substitution u = sin(x) which gives du = cos(x) dx. So, we obtain
∫
sec3 (x) dx
∫
du
=
(1 − u2 )2
∫
1
du
=
(u2 − 1)2
∫
1
=
du .
2
(u − 1) (u + 1)2
We look to find the partial fractions for the rational map
1
A
B
C
D
=
+
+
+
.
(u − 1)2 (u + 1)2
u − 1 (u − 1)2
u + 1 (u + 1)2
)
+C.
PRACTICE PROBLEMS: SET 6
11
We bring everything to a common denominator and therefore obtain
1 = A · (u − 1)(u + 1)2 + B · (u + 1)2 + C · (u − 1)2 (u + 1) + D · (u − 1)2 .
We plug in u = 1 and then u = −1 in the above identity of polynomials and obtain:
1 = B · 4; therefore B =
1
4
1 = D · 4; therefore D =
1
.
4
and
We substitute B = D =
1
4
in the above identity of polynomials and obtain:
1
= A · (u − 1)(u + 1)2 + B · (u + 1)2 + C · (u − 1)2 (u + 1) + D · (u − 1)2
1
1
= A · (u − 1)(u + 1)2 + · (u + 1)2 + C · (u − 1)2 (u + 1) + · (u − 1)2
4
4
1
1
2
2
2
= A(u − 1)(u + 2u + 1) + · (u + 2u + 1) + C · (u − 2u + 1)(u + 1) + · (u2 − 2u + 1)
4
4
u2
u 1
u2
u 1
3
2
3
2
= Au + Au − Au − A +
+ + + Cu − Cu − Cu + C +
− +
4) 2 4
4
2 4
(
1
1
= (A + C)u3 + A − C +
u2 − (A + C)u − A + C + .
2
2
Equating each term from each side of the above identity of polynomials, i.e. noting
that the coefficients of u3 , u2 , u and the constant term must be the same in both
the left and the right hand side above, we obtain

0 =
A+C



0 =
A − C + 12
−A − C
 0 =


1 = −A + C + 21 .
The first and the third equations yield
A = −C,
while the second and the fourth equations yield
1
A=C− .
2
Overall, we obtain that −C = C − 21 , and so,
C=
1
1
and thus, A = − .
4
4
We conclude that
1
1
1
− 41
1
4
4
4
=
+
+
+
.
(u − 1)2 (u + 1)2
u − 1 (u − 1)2
u + 1 (u + 1)2
12
MATH 101: PROF. DRAGOS GHIOCA
Therefore
∫
sec3 (x) dx
=
=
=
=
=
=
=
=
)
∫
∫
du
du
du
+
+
(u − 1)2
u+1
(u + 1)2
(
)
1
1
· − ln |u − 1| −
+ ln |u + 1| −
+C
u−1
u+1
(
)
1
1
· − ln | sin(x) − 1| −
+ ln | sin(x) + 1| −
+C
sin(x) − 1
sin(x) + 1
( )
sin(x) + 1 − 2 sin(x)
· ln +C
sin(x) − 1 sin2 (x) − 1
(sin(x) + 1)2 1 2 sin(x)
1
− ·
· ln +C
4
sin2 (x) − 1 4 − cos2 (x)
(sin(x) + 1)2 1 sin(x)
1
+ ·
· ln +C
4
− cos2 (x) 2 cos2 (x)
(
)2
1
sin(x) + 1
1
· ln
+ · tan(x) sec(x) + C
4
cos(x)
2
ln | tan(x) + sec(x)| + tan(x) sec(x)
+C.
2
1
4
1
4
1
4
1
4
(∫
·
−
du
+
u−1
∫
Problem 10. First we have to rationalize this fraction, i.e. we eliminate
the cube root.
We substitute x = u3 and thus dx = 3u2 du .
The above substitution may be viewed as a backward substitution just
as in the case of trigonometric substitutions.
Alternatively, we could have written the substitution as
u=
√
3
x,
but then
du =
1 −2
x 3 dx
3
is harder to express in the original integral. With a bit of algebra, we would’ve
obtained that
1 1
du = · 2 · dx,
3 u
and thus we would’ve been back to
3u2 du = dx .
Note that it is so much easier to have expressed x in terms of u in the
substitution than the other way around, i.e., in this case, the backward
substitution is much easier than the regular substitution.
PRACTICE PROBLEMS: SET 6
So, we obtain
∫
√
3
x
dx
x−1
13
∫
=
=
=
=
=
u
· 3u2 du
u3 − 1
∫
u3
du
3
u3 − 1
∫
1
3 1+ 3
du
u −1
∫
1
3u + 3
du
u3 − 1
∫
√
1
33x+3
du .
u3 − 1
In the above computation we also used a (very simple) long division computation
when we divided u3 by u3 − 1.
We only need to evaluate the last integral; for this we use the method of partial
fractions. We note that u3 − 1 = (u − 1)(u2 + u + 1), and so,
1
A
Bu + C
=
+
.
u3 − 1
u − 1 u2 + u + 1
In order to solve for A, B and C we bring to a common denominator and equate:
1 = A · (u2 + u + 1) + (Bu + C)(u − 1).
We plug in u = 1 in the above equation and get
1 = A · 3, and so, A =
Now we substitute A =
1
3
1
.
3
in the above equation and get
1
1
· (u2 + u + 1) + (Bu + C)(u − 1)
3
u 1
u2
+ + + Bu2 + (C − B)u − C
=
3
3)3
(
(
)
1
1
1
2
=
+B u +
+ C − B u + − C.
3
3
3
=
Therefore, equating the coefficients
identity, we obtain the system:

 0
0

1
of each power of u on both sides of the above
=
=
=
1
3
1
3
+B
+C −B
1
3 −C
The first equation yields B = − 13 , while the third equation yields C = − 23 . We
also note that substituting B = − 13 and C = − 32 in the second equation
yields an identity; this is important since it shows us that we are on the
right track with our solution.
14
MATH 101: PROF. DRAGOS GHIOCA
So,
∫
1
du
u3 − 1
∫
∫
− 13 u − 23
1
1
du +
du
3
u−1
u2 + u + 1
∫
1
1
u+2
=
ln |u − 1| −
du .
2
3
3
u +u+1
The last integral we split into two parts since (eventually) we want to use the
substitution
v = u2 + u + 1
for which
(
)
1
dv = (2u + 1) du = 2 · u +
du .
2
So,
∫
u+2
du
2
u +u+1
∫
∫
3
u + 12
2
=
du
+
du
u2 + u + 1
u2 + u + 1
∫ 1
∫
3
1
2 dv
=
+
du
2
v
2
u +u+1
∫
3
1
1
ln |v| +
du
=
2
2
(u + 12 )2 + 43
∫
1
3 4
1
2
=
ln(u + u + 1) + ·
du .
(
)2
2
2 3
2(u+ 12 )
√
+
1
3
=
In the last integral we make the substitution
v=
Therefore
∫
1
(
)2
2u+1
√
3
2(u + 21 )
2u + 1
2
√
= √
and so, dv = √ du .
3
3
3
∫
du =
+1
Therefore
∫ √
3
x
dx
x−1
∫
√
√
√
(
)
1
3
3
3
2u + 1
√
·
dv =
arctan(v)+C =
arctan
+C .
v2 + 1 2
2
2
3
3u3
du
−1
)
( √
) √
√
√
√
1 (√
23x+1
3
√
= 3 3 x + ln | 3 x − 1| − ln
+C.
x2 + 3 x + 1 − 3 arctan
2
3
√
Problem 11. We first do the substitution: u = x which yields u = x2 . So,
du = 2x dx. Then
∫
∫
√
sin( x) dx = sin(u) · 2u du .
=
u3
PRACTICE PROBLEMS: SET 6
15
Then we integrate by parts:
∫
∫
∫
2u sin(u) du = 2u·(− cos(u))′ du = −2u cos(u)− −2 cos(u) du = −2u cos(u)+2 sin(u)+C .
So,
∫
√
√
√
√
sin( x) dx = −2u cos(u) + 2 sin(u) + C = −2 x cos( x) + 2 sin( x) + C .
Problem 12. Either by using long division, or simply noting algebraically, we
have
(x2 + x + 1) − x
x
x2 + 1
=
=1− 2
.
x2 + x + 1
x2 + x + 1
x +x+1
So,
∫
∫
∫
x2 + 1
x
x
dx
=
1
−
dx
=
x
−
dx .
x2 + x + 1
x2 + x + 1
x2 + x + 1
Then we express the numerator x in terms of the derivative of the denominator
x2 + x + 1. So, we compare x with 2x + 1, and note that
1
1
x = · (2x + 1) − .
2
2
So,
∫ 1
∫
∫
∫
1
1
2x + 1
1
1
x
2 · (2x + 1) − 2
dx =
dx = ·
dx − ·
dx .
x2 + x + 1
x2 + x + 1
2
x2 + x + 1
2
x2 + x + 1
Doing a simple substitution u = x2 + x + 1 which yields du = (2x + 1) dx allows us
to compute
∫
∫
2x + 1
1
dx
=
du = ln |u| + C = ln(x2 + x + 1) + C .
x2 + x + 1
u
(Note that x2 + x + 1 = (x + 1/2)2 + 3/4 is always positive, which allows us to
remove the absolute values above.)
We are left to evaluate
∫
∫
1
1
dx
=
dx .
(
)
1 2
x2 + x + 1
x + 2 + 34
√
√
After making the substitution x + 12 = 23 · u, which yields dx = 23 · du, we have
√ ∫
√
(
)
∫
∫
3
1
1
3
1
2
2
2x + 1
2
√
√
√
dx
=
·
du
=
·
du
=
·arctan(u)+C
=
arctan
+C .
(
)
3
2
3
3
2
2
u2 + 1
3
3
3
x + 12 + 34
4 ·u + 4
4
In conclusion,
∫
(
)
2x+1
√
arctan
ln(x + x + 1)
x +1
3
√
dx = x −
+
+C.
x2 + x + 1
2
3
2
2