Review of Calculus 1 for Calculus 2 Students.
1.
Find y 0
y=
ex
4 + ex2
Solution:
Remembering the quotient rule and our rule for differentiating the exponential we begin:
d N
DN 0 − N D0
=
dx D
D2
2
y0 =
d f (x)
e
= ef (x) f 0 (x)
dx
2
2
(4 + ex )ex − ex ex 2x
ex
=
2
(4 + ex )2
+x
(1 − 2x) + 4ex
(4 + ex2 )2
2.
Find y 0
y = x arctan(x2 ) − ln(1 + x4 )
Solution:
Remembering our rules to differentiate products, arctan and logs:
d
(F S) = F S 0 +SF 0
dx
y0 = x ·
3.
Find y 0
d
f 0 (x)
arctan(f (x)) =
dx
1 + (f (x))2
d
f 0 (x)
ln(f (x)) =
dx
f (x)
2x
4x3
2x2 + (1 + x4 ) arctan(x2 ) − 4x3
+ arctan(x2 ) −
=
4
4
1+x
1+x
1 + x4
y = ex arcsec(ex ) −
p
1 − e2x
Solution:
The rule to differentiate products and the exponential function have already
been given so all we need is a rule to differentiate the arcsec function and the
generalized power rule.
d
f 0 (x)
p
arcsec(f (x)) =
dx
f (x) 1 − (f (x))2
y 0 = ex
√
ex
√
d
(f (x))n = n(f (x))n−1 f 0 (x)
dx
−1
1
ex
+ ex arcsec(ex ) − (1 − e2x ) 2 (−e2x 2) =
2x
2
1−e
e2x
ex + e2x
ex
+ ex arcsec(ex ) + √
=√
+ ex arcsec(ex )
1 − e2x
1 − e2x
1 − e2x
4.
Find y 0
arcsin(x2 )
y= √
1 − x4
Solution:
The only new formula we will need is a formula to differentiate the arcsin
function.
f 0 (x)
d
arcsin(f (x)) = p
dx
1 − (f (x))2
√
y0 =
2x
1 − x4 √1−x
− arcsin(x2 ) 21 (1 − x4 )
4
1 − x4
−1
2
4x3
2x −
=
After multiplying both numerator and denominator by
0
y =
√
2x 1 − x4 − 2x3 arcsin(x2 )
3
(1 − x4 ) 2
√
2x3 arcsin(x2 )
√
1−x4
1 − x4
1 − x4 gives:
5.
Find y 0
y = (x2 + 5) sin(1 + x2 ) − tan(1 + x2 )
Solution:
For this problem we will need differentiation formulas for the sin and tan
functions.
d
sin(f (x)) = cos(f (x))f 0 (x)
dx
d
tan(f (x)) = sec2 (f (x))f 0 (x)
dx
y 0 = (x2 + 5) cos(1 + x2 )2x + sin(1 + x2 )2x − sec2 (1 + x2 )2x
6.
Find the equation of the tangent line to f when x = 0.
f (x) = (1 + cos(x))sec(x)
Solution:
In order to find the equation of a tangent line we will need a point on the
line and the line’s slope at that point. We are given the x coordinate of our
point so all we need is the y coordinate and the slope. The y coordinate when
x = 0 is y = f (0) = 20 = 1; so we have a point: (0, 1). The slope comes from
the derivative when x = 0. Seeing that f is a function raised to a function we
will first need to take natural log of both sides and use our properties of logs
before we find f 0 .
ln(f ) = ln((1 + cos(x))sec(x) )
ln(f ) = sec(x) ln(1 + cos(x))
Differentiating gives:
f0
− sin(x)
− tan(x)
= sec(x)
+ln(1+cos(x)) sec(x) tan(x) =
+ln(1+cos(x)) sec(x) tan(x)
f
1 + cos(x)
1 + cos(x)
Multiplying by f = (1 + cos(x))sec(x) gives:
0
f (x) = (1 + cos(x))
sec(x)
− tan(x)
+ ln(1 + cos(x)) sec(x) tan(x)
1 + cos(x)
The slope of the tangent line is f 0 (0) = 0 so the equation of the tangent line
is:
y=1
7.
Evaluate.
Z
(x − 2)(x − 1)
dx
x
Solution:
Since we do not have a product or quotient rule for integrals we begin by
multiplying out our product and dividing out our quotient.
Z
(x − 2)(x − 1)
dx =
x
Z Z
x2 − 3x + 2
dx =
x
2
x2
x−3+
dx =
− 3x + 2 ln |x| + C
x
2
8.
Evaluate.
Z
0
Solution:
π
4
tan2 (x)dx
Since we do not have an integral formula for tan2 (x) but we do have an
integral formula for sec2 (x) and they are related by the trig identity 1+tan2 (x) =
sec2 (x) we proceed as follows:
Z
π
4
Z
2
tan (x)dx =
0
π
4
(sec2 (x) − 1)dx =
0
π4
π
π
π
tan(x) − x = tan
− − tan(0) − 0 = 1 −
4
4
4
0
9.
Evaluate.
Z
x4
x3 − 2x
dx
− 4x2 − 1
Solution:
Recognizing the degree of the numerator is one degree less that the degree
of the denominator we will begin by substituting for the denominator.
Z
x4
x3 − 2x
dx
− 4x2 − 1
u = x4 −4x2 −1
du = (4x3 −8x)dx
1
du = (x3 −2x)dx
4
Under this substitution our integral becomes
1
4
Z
1
1
du
= ln |u| + C = ln |x4 − 4x2 − 1| + C
u
4
4
10.
Evaluate.
Z p
1 + tan(x)
dx
cos2 (x)
Solution:
Here I see the derivative of 1 + tan(x) involved in the problem so I will
substitute for it.
Z p
1 + tan(x)
dx
cos2 (x)
u = 1 + tan(x)
du = sec2 (x)dx =
dx
cos2 (x)
Under this substitution our integral becomes
Z
1
u 2 du =
3
2 3
2
u 2 + C = (1 + tan(x)) 2 + C
3
3
11.
Evaluate.
Z
√
3
dx
−x2 + 14x − 33
Solution:
When integrating a constant over a square root of a quadratic start by
completing the square on the quadratic.
−x2 + 14x − 33 = −(x2 − 14x) − 33 = −(x2 − 14x + 49 − 49) − 33 =
− (x − 7) − 49 − 33 = −(x − 7)2 + 49 − 33 = −(x − 7)2 + 16
Z
2
3
√
dx =
2
−x + 14x − 33
Z
3
√
Z
3
p
du
= 3 arcsin
2
4 − u2
42
− (x − 7)2
Evaluate.
Solution:
u=x−7
du = dx
u
x−7
+ C = 3 arcsin
+C
4
4
12.
Z
dx
e2x
dx
4 + e4x
This integral looks closest to our arctan formula.
1
du = e2x dx
2
Under this substitution our integral becomes
u = e2x
1
2
Z
du = 2e2x dx
1 1
du
= · arctan
22 + u2
2 2
2x u
1
e
+ C = arctan
+C
2
4
2
13.
Evaluate.
Z
cos(2x)
dx
sin(x) − cos(x)
Solution:
Using the trig identity cos(2x) = cos2 (x) − sin2 (x)
Z
cos(2x)
dx =
sin(x) − cos(x)
Z
cos2 (x) − sin2 (x)
dx =
sin(x) − cos(x)
Z
(cos(x) − sin(x))(cos(x) + sin(x))
dx
−(cos(x) − sin(x))
Z
−
(cos(x) + sin(x))dx = −(sin(x) − cos(x)) + C
14.
Evaluate.
Z
e3x − e−3x
dx
e3x + e−3x
Solution:
u = e3x + e−3x
Z
du = (3e3x − 3e−3x )dx
1
e3x − e−3x
dx =
e3x + e−3x
3
Z
1
du = (e3x − e−3x )dx
3
du
1
1
= ln |u| + C = ln |e3x + e−3x | + C
u
3
3
15.
Evaluate.
Z
x3
dx
4 + x8
Solution:
u = x4
Z
1
x3
dx =
8
4+x
4
Z
1
du = x3 dx
4
du = 4x3 dx
du
1 1
= · arctan
2
2
2 +u
4 2
4
u
1
x
+ C = arctan
+C
2
8
2
16.
Evaluate.
Z
(x − 5)
√
1
dx
− 10x − 11
x2
Solution:
Completing the square on the contents of the square root gives:
x2 − 10x − 11 = x2 − 10x + 25 − 25 − 11 = (x − 5)2 − 36
Z
1
dx =
(x − 5) x2 − 10x − 11
√
u=x−5
Z
1
p
dx
(x − 5) (x − 5)2 − 62
du = dx
Z
1
1
√
dx =
2
u u − 62
(x − 5) (x −
−
1
u
1
x−5
arcsec
+ C = arcsec
+C
6
6
6
6
p
17.
Z
5)2
62
dx =
Evaluate.
Z
6x + 3
dx
x2 + 6x + 45
Solution:
To integrate a linear equation divided by a non-factorable quadratic start
by substituting for the quadratic.
u = x2 + 6x + 45
du = (2x + 6)dx
Seeing that we have a 6x in the numerator I will manipulate du so it too has
a 6x.
3du = (6x + 18)dx
Were this integral to be:
Z
6x + 18
dx
x2 + 6x + 45
we would have a useful substitution and the integral would be easy. The
solution to this problem is to force a 6x + 18 into the numerator by adding and
subtracting 15 in the numerator.
Z
6x + 3
dx =
x2 + 6x + 45
Z
6x + 3 + 15 − 15
dx =
x2 + 6x + 45
Z
6x + 18 − 15
dx
x2 + 6x + 45
Now I will separate this integral into two integrals: the first will the part
that we can handle with our substitution and the second will be whatever is
leftover.
Z
6x + 18 − 15
dx =
x2 + 6x + 45
Z
6x + 18
dx +
2
x + 6x + 45
Z
x2
−15
dx
+ 6x + 45
The first integral we have a substitution for and for the second integral, a
linear equation divided by a non-factorable quadratic, I will complete the square
on the quadratic.
x2 + 6x + 45 = x2 + 6x + 9 − 9 + 45 = (x + 3)2 + 36
Z
6x + 18
dx +
2
x + 6x + 45
1
3 ln |u| − 15 · arctan
6
Z
−15
dx = 3
2
x + 6x + 45
Z
du
− 15
u
Z
62
dx
=
+ (x + 3)2
x+3
5
x+3
2
+ C = 3 ln |x + 6x + 45| − arctan
+C
6
2
6
18.
Evaluate.
Z
2x3 + 6x2 + 8x + 27
dx
x2 + 4
Solution:
To integrate a rational function with the degree of the numerator greater
than or equal to the degree of the denominator always start with long division.
2x + 6
x2 + 4
2x3 + 6x2 + 8x + 27
− 2x3
Z
− 8x
6x2
+ 27
− 6x2
− 24
3
Z 3
2x3 + 6x2 + 8x + 27
dx =
2x + 6 + 2
dx =
x2 + 4
x +4
1
x
2
x + 6x + 3 · arctan
+C
2
2
19.
Evaluate.
Z
1
√ dx
1+ x
Solution:
Substituting for the denominator gives:
u=1+
√
dx
du = √
2 x
x
Seeing all we have in the numerator is just a dx we need to solve for dx in
terms of u and du alone.
u−1=
√
x
2(u − 1)du = dx
Under this substitution our integral becomes
Z
1
√ dx =
1+ x
2(u − ln |u|) + C = 2(1 +
√
2(u − 1)du
=2
u
Z
x − ln |1 +
√
Z 1−
1
du =
u
√
√
x|) + C = 2( x − ln |1 + x|) + C
20.
Evaluate.
Z
1
dx
√ √
x 1−x
Solution:
Trying to force this into the arcsin integral we must treat 1 − x as the a2 − u2
we see in our arcsin formula. So we make the substitution:
u=
Z
√
x
1
dx = 2
√ √
x 1−x
Z
dx
du = √
2 x
√
du
= 2 arcsin(u) + C = 2 arcsin( x) + C
1 − u2
21.
Evaluate.
Z
Solution:
dx
2du = √
x
e4x − 1
dx
e4x + 1
Seeing the similarity between this integral and problem 21 above I will start
be multiplying both numerator and denominator by e−2x .
e4x − 1
dx =
e4x + 1
Z
u = e2x + e−2x
e4x − 1 e−2x
dx =
·
e4x + 1 e−2x
Z
Z
e2x − e−2x
dx
e2x + e−2x
1
du = (e2x − e−2x )dx
2
du = (2e2x − 2e−2x )dx
1
e2x − e−2x
dx =
e2x + e−2x
2
Z
Z
du
1
1
= ln |u| + C = ln |e2x + e−2x | + C
u
2
2
22.
Evaluate.
Z
(9x2 + 4x) ln(x)dx
Solution:
To integrate a polynomial times a log always use Integration By Parts (IBP).
Remember the integration by parts formula says:
Z
Z
udv = vu −
u = ln(x)
du =
dv = (9x2 + 4x)dx
Z
2
3
vdu
dx
x
v = 3x3 + 2x2
2
(9x + 4x) ln(x)dx = (3x + 2x ) ln(x) −
3
2
(3x + 2x ) ln(x) −
23.
Z
Z
(3x3 + 2x2 )
dx
=
x
(3x2 + 2x)dx = (3x3 + 2x2 ) ln(x) − (x3 + x2 ) + C
Evaluate.
Z
ex sin(2x)dx
Solution:
This integral is a cyclic IBP problem. Meaning that after some number of
uses of IBP we will end up with the same integral we started with.
dv = ex dx
u = sin(2x)
v = ex
du = 2 cos(2x)dx
Z
ex sin(2x)dx = ex sin(2x) − 2
Z
ex cos(2x)dx
Applying IBP again on this last integral
dv = ex dx
u = cos(2x)
du = −2 sin(2x)dx
Z
v = ex
Z
x
x
e sin(2x)dx = e sin(2x) − 2 e cos(2x) + 2 e sin(2x)dx
x
Z
x
x
x
x
e sin(2x)dx = e sin(2x) − 2e cos(2x) − 4
Z
ex sin(2x)dx
Solving for the desired integral.
Z
5
Z
24.
Evaluate.
ex sin(2x)dx = ex sin(2x) − 2ex cos(2x)
1 x
x
e sin(2x) − 2e cos(2x) + C
e sin(2x)dx =
5
x
Z
2x arctan(x)dx
Solution:
To integrate a polynomial times an inverse trig function always start with
IBP.
u = arctan(x)
du =
Z
dv = 2xdx
1
dx
1 + x2
v = x2
2x arctan(x)dx = x2 arctan(x) −
Z
x2
dx
1 + x2
We now need to integrate a rational function whose degree of the numerator
is equal to the degree of the denominator so we attack with long division.
1
x2 + 1
x2
− x2 − 1
Z
−1
Z 2
2x arctan(x)dx = x arctan(x) −
1−
1
dx =
1 + x2
x2 arctan(x) − x + arctan(x) + C
25.
Evaluate.
Z
ln(9 + x2 )dx
Solution:
u = ln(9 + x2 )
dv = dx
du =
Z
2x
dx
9 + x2
v=x
ln(9 + x2 )dx = x ln(9 + x2 ) −
Z
2x2
dx
9 + x2
We now need to integrate a rational function whose degree of the numerator
is equal to the degree of the denominator so we attack with long division.
2
2
x +9
2x
2
− 2x2 − 18
− 18
Z
Z ln(9 + x2 )dx = x ln(9 + x2 ) −
2−
Z
18
dx =
9 + x2
x
1
ln(9 + x2 )dx = x ln(9 + x2 ) − 2x − 18 · arctan
+C =
3
3
x ln(9 + x2 ) − 2x + 6 arctan
x
+C
3
26.
Z
cos3 (x) sin2 (x)dx
Solution:
Swapping out a cos2 (x) with a 1 − sin2 (x) gives:
Z
cos3 (x) sin2 (x)dx =
Z
(1 − sin2 (x))(x) sin2 (x) cos(x)dx
Now we have a simple substitution problem
u = sin(x)
Under this substitution we get
du = cos(x)dx
Z
(1 − sin2 (x))(x) sin2 (x) cos(x)dx =
Z
Z
(1 − u2 )u2 du =
(u2 − u4 )du =
u5
sin3 (x) sin5 (x)
u3
−
+C =
−
+C
3
5
3
5
27.
Z
sec4 (x) tan3 (x)dx
Solution:
Swapping out a sec2 (x) for a 1 + tan2 (x) gives:
Z
sec4 (x) tan3 (x)dx =
Z
(1 + tan2 (x)) tan3 (x) sec2 (x)dx
du = sec2 (x)dx
u = tan(x)
Z
(1 + tan2 (x)) tan3 (x) sec2 (x)dx =
Z
(1 + u2 )u3 du =
Z
(u5 + u3 )du =
u6
u4
tan6 (x) tan4 (x)
+
+C =
+
+C
6
4
6
4
28.
Z
sec4 (x) tan6 (x)dx
Solution:
Swapping out a sec2 (x) for a 1 + tan2 (x) gives:
Z
sec4 (x) tan6 (x)dx =
u = tan(x)
Z
(1 + tan2 (x)) tan6 (x) sec2 (x)dx
du = sec2 (x)dx
Z
(1 + tan2 (x)) tan6 (x) sec2 (x)dx =
Z
(1 + u2 )u6 du =
Z
(u8 + u6 )du =
u9
u7
tan9 (x) tan7 (x)
+
+C =
+
+C
9
7
9
7
29.
Z
sin2 (x) cos2 (x)dx
Solution:
To integrate even powers of sin and cosine we must use our power reducing
formulas:
sin2 (x) =
Z
2
1
(1 − cos(2x))
2
2
Z
sin (x) cos (x)dx =
1
4
Z
cos2 (x) =
1
1
(1 − cos(2x)) (1 + cos(2x))dx =
2
2
1
(1 − cos (2x))dx =
4
2
1
(1 + cos(2x))
2
Z
sin2 (2x)dx
To integrate even powers of sin and cosine we must use our power reducing
formulas
1
4
Z
1
1
sin(4x)
(1 − cos(4x))dx =
x−
+C
2
8
4
30.
Z
tan3 (x)
dx
cos3 (x)
Solution:
Converting this integral into sin and cosine gives:
Z
tan3 (x)
dx =
cos3 (x)
sin3 (x)
1
dx
cos3 (x) cos3 (x)
Z
Swapping out a sin2 (x) with a 1 − cos2 (x) gives:
Z
sin3 (x)
dx =
cos6 (x)
(1 − cos2 (x)) sin(x)
dx
cos6 (x)
du = − sin(x)dx
u = cos(x)
Z
Z
(1 − cos2 (x)) sin(x)
dx = −
cos6 (x)
− du = sin(x)dx
1 − u2
du =
u6
Z
Z
(u−4 − u−6 )du =
u−3
u−5
1
1
−
+C =
−
+C
−3
−5
5 cos5 (x) 3 cos3 (x)
31.
Z
1
3
(4 − x2 ) 2
dx.
Solution:
Here I will use a substitution to force a trig identity onto this integral.
x = 2 sin θ
Z
1
Z
3 dx =
(4 − x2 ) 2
Z
Z
1
(4 cos2
θ)
3
2
dx = 2 cos θdθ
2 cos θdθ =
1
3
(4 − 4 sin2 θ) 2
2 cos θdθ
1
=
3
8 cos θ
4
Z
2 cos θdθ =
sec2 θdθ =
1
tan θ + C
4
Now I must convert back to the variable x from the variable θ. To do this I
will draw a triangle based on my original substitution.
x = 2 sin θ
sin θ =
x
opposite
=
2
hypotenuse
2
θ
√
x
4 − x2
From the triangle we see that tan θ =
Z
1
3
(4 − x2 ) 2
dx =
√ x
.
4−x2
So
1
1
x
+C
tan θ + C = √
4
4 4 − x2
32.
Z
2
dx
x 9 − x2
√
Solution:
Here I will use a substitution to force a trig identity onto this integral.
x = 3 sin θ
Z
2
dx =
x 9 − x2
√
Z
Z
2
3 sin θ
2
3
dx = 3 cos θdθ
p
2
3 cos θdθ = 2
9 − 9 sin θ
Z
csc θdθ =
cos θ
√
dθ =
sin θ 9 cos2 θ
−2
ln | csc θ + cot θ| + C
3
Now I must convert back to the variable x from the variable θ. To do this I
will draw a triangle based on my original substitution.
x = 3 sin θ
3
θ
√
9 − x2
x
sin θ =
x
opposite
=
3
hypotenuse
From the triangle we see that csc θ =
√
3
x
and cot θ =
9−x2
x
making our
solution:
−2
3
−2
ln | csc θ + cot θ| + C =
ln | +
3
3
x
√
9 − x2
|+C
x
33.
Z
x2
√
1
dx.
9 + x2
Solution:
Here I will use a substitution to force a trig identity onto this integral.
dx = 3 sec2 θdθ
x = 3 tan θ
Z
Z
1
1
√
√
3 sec2 θdθ =
dx =
2
2
2
x 9+x
9 tan θ 9 + 9 tan2 θ
Z
Z
Z
1
1
sec2 θdθ
1
sec2 θdθ
sec θdθ
√
=
=
2
2
2
3
3
9
tan
θ3
sec
θ
tan2 θ
tan θ 9 sec θ
Converting to sin and cosine gives:
1
9
Z
1
cos2 θ
1
·
dθ =
2
cos θ sin θ
9
u = sin θ
1
9
Z
cos θ
1
2 dθ = 9
sin θ
Z
cos θ
dθ
sin2 θ
du = cos θdθ
Z
u−2 du =
1 u−1
+C =
9 −1
1 −1
1 −1
−1
·
+C = ·
+C =
csc θ + C
9 u
9 sin θ
9
Now I must convert back to the variable x from the variable θ. To do this I
will draw a triangle based on my original substitution.
x = 3 tan θ
tan θ =
x
opposite
=
3
adjacent
√
9 + x2
x
θ
3
√
From the triangle we see that csc θ =
9+x2
x
−1
−1
csc θ + C =
·
9
9
34.
Z √
√
and our solution is:
9 + x2
+C
x
x2 − 4
dx.
x
Solution:
Here I will use a substitution to force a trig identity onto this integral.
x = 2 sec θ
Z √
dx = 2 sec θ tan θdθ
4 sec2 θ − 4
2 sec θ tan θdθ =
2 sec θ
Z
2
Z p
2
4 tan θ tan θdθ = 2
Z
tan2 θdθ
(sec2 θ − 1)dθ = 2(tan θ − θ) + C
Now I must convert back to the variable x from the variable θ. To do this I
will draw a triangle based on my original substitution.
x = 2 sec θ
√
x
sec θ =
x2 − 4
θ
2
Using the triangle to convert back to x
x
hypotenuse
=
2
adjacent
√ 2
x −4
x
2(tan θ − θ) + C = 2
− arcsec
+C
2
2
35.
Z
−3x2 + 8x + 9
dx.
x3 − 6x2 + 9x
Solution:
We will start by factoring the denominator and then we will apply partial
fractions.
−3x2 + 8x + 9
−3x2 + 8x + 9
−3x2 + 8x + 9
=
=
3
2
2
x − 6x + 9x
x(x − 6x + 9)
x(x − 3)2
Applying partial fractions:
A
B
C
−3x2 + 8x + 9
= +
+
x(x − 3)2
x
x − 3 (x − 3)2
Multiplying both sides by the least common denominator: x(x − 3)2
−3x2 + 8x + 9
A
B
C
= +
+
x(x − 3)2
x(x − 3)2
x
x − 3 (x − 3)2
−3x2 + 8x + 9 = A(x − 3)2 + Bx(x − 3) + Cx
If x = 0 this equation reduces to:
9 = 9A
A=1
If x = 3 this equation reduces to:
6 = 3C
C=2
If x = 1 this equation reduces to:
14 = 4A − 2B + C
8 = −2B
Using these values of A, B and C we obtain:
B = −4
Z
−3x2 + 8x + 9
dx =
x3 − 6x2 + 9x
Z ln |x| − 4 ln |x − 3| +
36.
Z −4
2
1
−4
1
−2
dx
=
+
+
+
+2(x−3)
dx =
x x − 3 (x − 3)2
x x−3
2(x − 3)−1
2
+ C = ln |x| − 4 ln |x − 3| −
+C
−1
x−3
Z
x3
−3x − 1
dx.
− 3x2 + x − 3
Solution:
We will start by factoring the denominator and then we will apply partial
fractions.
−3x − 1
−3x − 1
−3x − 1
= 2
=
x3 − 3x2 + x − 3
x (x − 3) + x − 3
(x − 3)(x2 + 1)
Applying partial fractions:
A
Bx + C
−3x − 1
=
+ 2
(x − 3)(x2 + 1)
x−3
x +1
Multiplying both sides by the least common denominator: (x − 3)(x2 + 1)
−3x − 1
A
Bx + C
=
+
(x − 3)(x2 + 1)
(x − 3)(x2 + 1)
x−3
x2 + 1
−3x − 1 = A(x2 + 1) + (Bx + C)(x − 3)
If x = 3 this equation becomes:
−10 = 10A
A = −1
If x = 0 this equation becomes:
−1 = A − 3C
0 = −3C
C=0
If x = 1 this equation becomes:
−4 = 2A − 2(B + C)
− 2 = −2B
B=1
Using these values of A, B and C we obtain:
Z
−3x − 1
dx =
3
x − 3x2 + x − 3
Z x
−1
+ 2
x−3 x +1
−1
dx +
x−3
Z
=
Z
x2
x
dx
+1
Using a substitution on the second integral gives:
u = x2 + 1
Z
−1
dx +
x−3
Z
x
dx =
2
x +1
1
−1
dx +
x−3
2
Z
− ln |x − 3| +
37.
1
du = xdx
2
du = 2xdx
Z
Z
1
du
= − ln |x − 3| + ln |u| + C =
u
2
1
ln |x2 + 1| + C
2
2x5 + 18x3 + 10x2 + 36
dx.
x4 + 9x2
Solution:
Since the degree of the numerator is greater than the degree of the denominator we start with long division.
2x
4
x + 9x
2
5
3
2
2x + 18x + 10x + 36
− 2x5 − 18x3
Z
2x5 + 18x3 + 10x2 + 36
dx =
x4 + 9x2
Z 2x+
Z 10x2 + 36
10x2 + 36
dx
=
2x+
dx
x4 + 9x2
x2 (x2 + 9)
Applying partial fractions:
10x2 + 36
Ax + B
Cx + D
=
+ 2
2
2
2
x (x + 9)
x
x +9
Multiplying both sides by the least common denominator: x2 (x2 + 9)
10x2 + 36
Ax + B
(Cx + D) 2 2
=
+
x (x + 9)
x2 (x2 + 9)
x2
x2 + 9
10x2 + 36 = (Ax + B)(x2 + 9) + (Cx + D)x2
If x = 0 this equation becomes:
36 = 9B
B=4
If x = 1 this equation becomes:
46 = 10(A + B) + C + D
6 = 10A + C + D
If x = −1 this equation becomes:
46 = 10(B − A) − C + D
6 = −10A − C + D
Adding the two equations 6 = 10A + C + D and 6 = −10A − C + D gives:
12 = 2D
D=6
Replacing D with 6 in the two equations 6 = 10A + C + D and 6 = −10A −
C + D gives:
10A + C = 0
C = −10A
If x = 2 we get:
76 = 13(2A+B)+4(2C +D)
0 = 26A+8C
A=0
Z 2x +
x2 −
0 = 26A+8(−10A)
C=0
Z 10x2 + 36
4
6
dx
=
2x
+
+
dx =
x2 (x2 + 9)
x2
x2 + 9
4
1
+ 6 · arctan
x
3
x
4
x
+ C = x2 − + 2 arctan
+C
3
x
3