CORRECTION TO SCHAUM’S OUTLINES: CALCULUS, FOURTH
EDITION
TJ LEONE
The following is a correction to EXAMPLE 9 on page 308 of Schaum’s Outlines: Calculus,
Fourth Edition, by Frank Ayres, Jr. and Elliot Mendelson. In the text, the partial fraction
decomposition is wrong, which means that most of the steps in the solution wrong. So
instead of trying to make line by line corrections, I’m just rewriting the whole solution.
The partial fraction decomposition in the book goes wrong in the “Multiply out on the
right” step. The book has the following:
x − 1 = (A + B + D)x4 + (B + E)x3 + (3A + C + D)x2 + (2C + E)x + 2A
The correct equation, show below is:
x − 1 = (A + B + D)x4 + (C + E)x3 + (3A + 2B + D)x2 + (2C + E)x + 2A
Here is the entire problem and correct solution:
Find
Z
x(x2
(x − 1)
dx
+ 1)(x2 + 2)
Represent the integrand as follows:
x(x2
(x − 1)
A Bx + C
Dx + E
= + 2
+ 2
2
+ 1)(x + 2)
x
x +1
x +2
Clear the denominators by multiplying by x(x2 + 1)(x2 + 2)
x − 1 = A(x2 + 1)(x2 + 2) + (Bx + C)x(x2 + 2) + (Dx + E)x(x2 + 1)
1
2
TJ LEONE
Multiply out on the right:
x − 1 = (A + B + D)x4 + (C + E)x3 + (3A + 2B + D)x2 + (2C + E)x + 2A
Comparing coefficients, we get:
2A = −1,
2C + E = 1,
3A + 2B + D = 0,
C + E = 0,
A+B+D =0
Hence, A = − 12 and, therefore, 2B + D = 32 , B + D = 12 . From the latter two equations,
B = 1. From 2C + E = 1 and C + E = 0, we get C = 1. Then, from B + D = 12 , it follows
that D = − 21 . Finally, from C + E = 0, E = −1.
Thus,
x(x2
(x − 1)
11
x+1
1 x+2
=−
+ 2
−
2
+ 1)(x + 2)
2 x x + 1 2 x2 + 2
Hence,
Z
(x − 1)
1
dx = − ln x +
2
2
x(x + 1)(x + 2)
2
Z
x+1
1
dx −
2
x +1
2
Z
x+2
dx
x2 + 2
Now, if we let x = tan θ, then dx = sec2 θdθ, and
Z
Z
Z
x+1
tan θ + 1
2
dx =
sec θdθ = tan θ + 1dθ = ln | sec θ| + θ + C
x2 + 1
sec2 θ
√
√
Since x = tan θ, we have sec θ = tan2 θ + 1 = x2 + 1 and θ = tan−1 x, so
Z
p
1
x+1
dx
=
ln
x2 + 1 + tan−1 x + C = ln (x2 + 1) + tan−1 x + C
2
x +1
2
√
2
Also, if we let x = 2 tan φ, then dx = 2 sec φdφ, and
√
Z √
Z
√
x+2
1
2 tan φ + 2 √
1
1
2
2
dx = −
2 sec φdφ = −
tan φ + 2dφ = − ln | sec φ|−
φ+C
2
2
x +2
2
2 sec φ
2
2
2
p
√
√
√
Since x = 2 tan φ, we have 2 sec φ = 2 tan2 φ + 2 = x2 + 2 and φ = tan−1 √x2 , so
√
√
√
Z
x+2
1 2 2
1
2
x
−1 x
2
dx = − ln √
tan √ + C = − ln (x + 2) −
tan−1 √ + K
−
x2 + 2
2 x2 + 2 2
4
2
2
2
1
−
2
Z
CORRECTION TO SCHAUM’S OUTLINES: CALCULUS, FOURTH EDITION
3
where K = C − 14 ln 2
Therefore,
Z
√
2
(x − 1)
1
1
1
x
2
−1
2
dx = − ln x+ ln (x + 1)+tan x− ln (x + 2)−
tan−1 √ +C
2
2
x(x + 1)(x + 2)
2
2
4
2
2
4
TJ LEONE
Below is the SAGE notebook I used to verify my answer: