Question

Evaluate the integral $\int_{0}^{1} \frac{1 - 2 x}{(2 x + 1) {(x + 1)}^{2}} d x$

Accepted Answer

Let $I = \int_{0}^{1} \frac{1 - 2 x}{(2 x + 1) {(x + 1)}^{2}} d x$

Consider, $\frac{1 - 2 x}{(x + 1)^{2} (2 x + 1)} = \frac{a_{0}}{2 x + 1} + \frac{a_{1}}{x + 1} + \frac{a_{2}}{(x + 1)^{2}}$

$\Rightarrow 1 - 2 x = a_{0} (x + 1)^{2} + a_{1} (2 x + 1) (x + 1) + a_{2} (2 x + 1)$

$\Rightarrow 1 - 2 x = a_{0} (x^{2} + 1 + 2 x) + a_{1} (2 x^{2} + x + 2 x + 1) + 2 a_{2} x + a_{2}$

$\Rightarrow 1 - 2 x = a_{0} x^{2} + a_{0} + 2 a_{0} x + 2 a_{1} x^{2} + 3 a_{1} x + a_{1} + 2 a_{2} x + a_{2}$

$\Rightarrow 1 - 2 x = x^{2} (a_{0} + 2 a_{1}) + x (2 a_{0} + 3 a_{1} + 2 a_{2}) + a_{2} + a_{0} + a_{1}$

On comparing the coefficients of like powers of $x$ , we get:

$a_{0} + 2 a_{1} = 0, 2 a_{0} + 3 a_{1} + 2 a_{2} = - 2, a_{2} + a_{0} + a_{1} = 1$

On solving these, we get:

$a_{0} = 8, a_{2} = - 3, a_{1} = - 4$

Thus, $I = \int_{0}^{1} (\frac{8}{2 x + 1} - \frac{4}{x + 1} - \frac{3}{(x + 1)^{2}}) d x$

$\Rightarrow I = {[\frac{8 \ln |2 x + 1|}{2} - 4 \ln |x + 1| + \frac{3}{x + 1}]}_{0}^{1}$

$\Rightarrow I = [4 \ln |2 (1) + 1| - 4 \ln |1 + 1| + \frac{3}{1 + 1}] - [4 \ln |2 (0) + 1| - 4 \ln |0 + 1| + \frac{3}{0 + 1}]$

$\Rightarrow I = [4 \ln 3 - 4 \ln 2 + \frac{3}{2}] - [4 \ln 1 - 4 \ln 1 + 3]$

$\Rightarrow I = [\ln 3^{4} - \ln 2^{4} + \frac{3}{2}] - [0 - 0 + 3] [∵ \ln 1 = 0]$

$\Rightarrow I = \ln 81 - \ln 16 + \frac{3}{2} - 3$

$\Rightarrow I = \ln \frac{9^{2}}{4^{2}} - \frac{3}{2}$

$\Rightarrow I = 2 \ln \frac{9}{4} - \frac{3}{2}$

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 6: EXERCISE 8E

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 6: EXERCISE 8E

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 6: EXERCISE 8E with Hints & Solutions

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