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 fractiondecomposition is wrong, which means that most of the steps in the solution wrong. Soinstead 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 theright” 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 ∫(x− 1)

x(x2 + 1)(x2 + 2)dx

Represent the integrand as follows:

(x− 1)

x(x2 + 1)(x2 + 2)=A

x+Bx+ C

x2 + 1+Dx+ E

x2 + 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 = 3

2 , 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 = −12 . Finally, from C + E = 0, E = −1.

Thus,

(x− 1)

x(x2 + 1)(x2 + 2)= −1

2

1

x+

x+ 1

x2 + 1− 1

2

x+ 2

x2 + 2

Hence, ∫(x− 1)

x(x2 + 1)(x2 + 2)dx = −1

2lnx+

∫x+ 1

x2 + 1dx− 1

2

∫x+ 2

x2 + 2dx

Now, if we let x = tan θ, then dx = sec2 θdθ, and∫x+ 1

x2 + 1dx =

∫tan θ + 1

sec2 θsec2 θdθ =

∫tan θ + 1dθ = ln | sec θ|+ θ + C

Since x = tan θ, we have sec θ =√

tan2 θ + 1 =√x2 + 1 and θ = tan−1 x, so∫

x+ 1

x2 + 1dx = ln

√x2 + 1 + tan−1 x+ C =

1

2ln (x2 + 1) + tan−1 x+ C

Also, if we let x =√

2 tanφ, then dx = 2 sec2 φdφ, and

−1

2

∫x+ 2

x2 + 2dx = −1

2

∫ √2 tanφ+ 2

2 sec2 φ

√2 sec2 φdφ = −1

2

∫tanφ+

√2dφ = −1

2ln | secφ|−

√2

2φ+C

Since x =√

2 tanφ, we have√

2 secφ =√

2 tan2 φ+ 2 =√x2 + 2 and φ = tan−1 x√

2, so∫

x+ 2

x2 + 2dx = −1

2ln

∣∣∣∣∣√

2√x2 + 2

∣∣∣∣∣−√

2

2tan−1

x√2

+C = −1

4ln (x2 + 2)−

√2

2tan−1

x√2

+K

CORRECTION TO SCHAUM’S OUTLINES: CALCULUS, FOURTH EDITION 3

where K = C − 14 ln 2

Therefore,

∫(x− 1)

x(x2 + 1)(x2 + 2)dx = −1

2lnx+

1

2ln (x2 + 1)+tan−1 x−1

4ln (x2 + 2)−

√2

2tan−1

x√2

+C

4 TJ LEONE

Below is the SAGE notebook I used to verify my answer: