April 15, 2011
18.01 Problem Set 10 Solutions
Part II: 15 points
1) This problem is about
Z
x4
2xdx
.
+ x2 + 1
a) Compute the integral using first the “obvious” substitution u = x2 .
This gives
Z
du
=
u2 + u + 1
Z
du
= tan−1
(u + 1/2)2 + 3/4
u + 1/2
√
3/2
−1
= tan
2u + 1
√
3
−1
= tan
2x2 + 1
√
3
b) Factor the denominator as a product of two quadratic polynomials. (Hint: the factors
have integer coefficients, so you can find them by a little bit of guessing.)
One way to do this is to “complete the square,” adding and subtracting x2 :
x4 + x2 + 1 = x4 + 2x2 + 1 − x2 = (x2 + 1)2 − x2 = (x2 + 1 − x)(x2 + 1 + x) = (x2 + x + 1)(x2 − x + 1).
c) Compute the partial fraction expansion of the integrand, and use this to compute the
integral.
The expansion is
Ax + B
Cx + D
+ 2
.
2
x +x+1 x −x+1
Adding these two terms and equating with the integrand gives
(Ax + B)(x2 − x + 1) + (Cx + D)(x2 + x + 1) = 2x.
Equating coefficients of powers of x gives four equations
A + C = 0,
−A + B + C + D = 0,
A − B + C + D = 2,
B + D = 0.
These equations can be solved to give A = C = 0, D = −B = 1. That is, the expansion is
x4
2x
−1
1
= 2
+ 2
.
2
+x +1
x +x+1 x −x+1
The denominators on the right are (x + 1/2)2 + 3/4 and (x − 1/2)2 + 3/4, so the integral is
− tan
−1
2x + 1
√
3
+ tan
−1
2x − 1
√
3
.
d) Explain why your answer in (c) is the same as your answer in (a) (except perhaps an
additive constant). (I’m looking for something better than “I didn’t make a mistake in
either calculation, so they must give the same answer.”)
There is an addition formula for tangent:
tan(θ − φ) =
tan(θ) − tan(φ)
.
1 + tan θ tan φ
Write a = tan(θ), b = tan(φ), so that θ = tan−1 (a) and φ = tan−1 (b). Then the addition formula can
be read as
a−b
,
tan(tan−1 (a) − tan−1 (b)) =
1 + ab
or
a−b
tan−1 (a) − tan−1 (b) = tan−1 (
).
1 + ab
Plugging this into the antiderivative in (c) gives (with b =
tan−1
2x−1
√
3
1+
2x+1
√
3
2x+1
√
3
2
4x −1
3
−
and a =
2x−1
√ )
3
!
.
A little algebra converts this into
tan−1
−2
√
3
4x2 +2
3
!
= tan−1
√ !
− 3
2x2 + 1
The argument of tan−1 is now the negative inverse of the argument in (a). One more trig identity
tan(−π/2 + θ) =
can be read as
tan−1 (
−1
tan(θ)
−1
) = tan−1 (u) − π/2.
u
So the answer in (c) differs from the answer in (a) by subtracting the constant π/2. (Actually one should
be a bit more careful about where these inverse tangents take values; for example, a more careful version
of the last formula is
tan−1 (u) − π/2 if u > 0,
−1 −1
tan (
)=
tan−1 (u) + π/2 if u < 0.
u
But if this care is taken, it turns out that the answer in (a) is always greater by π/2 than the answer
in (c).