Question

Find : ∫((x² + 1)/(x - 1)²(x + 3)) dx

Answer 1

\(\frac{x^2 + 1}{(x - 1)^2 (x + 3)} = \frac A{(x - 1)^2} + \frac B{(x - 1)} + \frac C{(x + 3)}\)

\(= \frac{Ax + 3A + B[(x - 1)(x + 3)] + C(x - 1)^2}{(x - 1)^2 (x + 3)}\)

\((x^2 + 1) = Ax + 3A + 8x^2 - 3B + 2Bx + Cx^2 + C - 2Cx\)

Coeff. of x² = 1 = (B + C)    ......(1)

Coeff. of x = 0 = A + 2B - 2C    .....(2)

Coeff. of constant = 1 = 3A - 3B + C  ......(3)

(2) + (3),

4A - B - C = 1

Using eq. (1),

4A - 1(1) = 1

4A = 2

⇒ A = \(\frac 12\)

Put A in eq. (2),

A + 2B - 2C = 0

(B - C) = \(- (\frac 12). \frac 12\)

B - C = \(- \frac 14\)

B + C = 1

2B = \( \frac 34\)

⇒ B = \( \frac 38\)

C = \( \frac58\)

⇒ \(\int \frac{(x^2 + 1)}{(x - 1)^2 (x + 3)^2} dx = \int \frac A {(x - 1)^2} dx+ \int \frac{B}{x - 1} dx + \int \frac{C}{x + 3} dx\)

\(= \frac 12 \int \frac 1{(x - 1)^2} dx + \frac 38 \int \frac 1{x - 1} dx + \frac 58 \int \frac 1{x + 3} dx\)

\( (x - 1) = t\)

\(dx = dt \)

\(= \frac 12 \int \frac 1{t^2} dt + \frac {3}{8} \log|x - 1| + \frac 58 \log|x + 3| + C\)

\(= \frac 12 \int {t^{-2}} dt + \frac {3}{8} \log|x - 1| + \frac 58 \log|x + 3| + C\)

\(= \frac 12 \int \frac{{t^{-2+1}}}{-2 + 1} + \frac {3}{8} \log|x - 1| + \frac 58 \log|x + 3| + C\)

\(= -\frac 12 \int \frac{1}{x-1} + \frac {3}{8} \log|x - 1| + \frac 58 \log|x + 3| + C\)

Answer 2

Given integral

By putting x =1, we get B=1/2,

By putting x = -3, we get C = 10/16 = 5/8 and

By putting x = 2, we get 5A + 5B + C = 5

⇒ 5A = 5 - 5/8 - 5/2

⇒ 5A = (40-5-20)/8 = 15/8

⇒ A =15/(8 x 5) = 3/8