Question

Evaluate:

\int \frac{3 x + 1}{(x + 2) (x - 2)^{2}} d x

Accepted Answer

Given, $\int \frac{3 x + 1}{(x + 2) (x - 2)^{2}} d x$

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

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

$\Rightarrow (3 x + 1) = A \times (x^{2} - 4 x + 4) + B \times (x^{2} - 4) + C \times (x + 2)$

$\Rightarrow (3 x + 1) = A x^{2} - 4 A x + 4 A + B x^{2} - 4 B + C x + 2 C$

$\Rightarrow (3 x + 1) = (A + B) x^{2} + (- 4 A + C) x + (4 A - 4 B + 2 C)$

Compare the corresponding coefficients of powers of $x$ , we get

$\begin{array}{l} A + B = 0 \end{array}$

$\Rightarrow - 4 A + C = 3$

$\Rightarrow 4 A - 4 B + 2 C = 1$

Solving using elimination method, we get

$A = - \frac{5}{16}, B = \frac{5}{16}, C = \frac{7}{4}$

$∴ \frac{((3 x + 1))}{(x + 2) (x - 2)^{2}} = \frac{- \frac{5}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{7}{4}}{(x - 2)^{2}}$

$\Rightarrow \int \frac{3 x + 1}{(x + 2) (x - 2)^{2}} d x = - \frac{5}{16} \int \frac{1}{x + 2} d x + \frac{5}{16} \int \frac{1}{x - 2} d x + \frac{7}{4} \int \frac{1}{(x - 2)^{2}} d x$

$\Rightarrow \int \frac{3 x + 1}{(x + 2) (x - 2)^{2}} d x = \frac{- 5}{16} \log |x + 2| + \frac{5}{16} \log |x - 2| - \frac{7}{4 (x - 2)} + C^{'}$

R S Aggarwal Solutions for Exercise 1: EXERCISE

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