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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Prove that $\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
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Answer

Given integral is: $\int_{1}^{3} \frac{d x}{x^{2} (x+1)}$ 
To Prove: $\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$ 
Let $I=\frac{d x}{\left(x^{2}\right)(x+1)}$ 
Using partial fraction:
Let $\frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1} .....(i)$
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}$ = $\frac{A(x)(x+1)+B(x+1)+C\left(x^{2}\right)}{(x+1)\left(x^{2}\right)}$ 
$\Rightarrow 1=\mathrm{A}\left(\mathrm{x}^{2}+\mathrm{x}\right)+(\mathrm{Bx}+\mathrm{B})+\mathrm{Cx}^{2}$ 
$\Rightarrow 1 = Ax^2 + Ax + B + Bx + Cx^2$
$\Rightarrow 1 = B + (A + B)x + (A + C)x^{2 }$
Equating the coefficients of $x, x^2 $ and constant value. We get:
$B = 1$
$A + B = 0 \Rightarrow A = -B \Rightarrow A = -1$
$A + C = 0 \Rightarrow C = -A \Rightarrow C = 1$
Put these values in equation $(i)$
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$ 
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}$ 
$\Rightarrow \int \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\int-\frac{1}{x} d x+\int \frac{1}{\left(x^{2}\right)} d x+\int \frac{1}{(x+1)} d x$ 
$\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\left[-\log |x|-x^{-1}+\log |x+1|\right]_{1}^{3}$ 
$\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\left[-\frac{1}{x}+\log \left|\frac{x+1}{x}\right|\right]_{1}^{3}$ 
= $\left[-\frac{1}{3}+\log \left|\frac{3+1}{3}\right|-\left(-\frac{1}{1}+\log \left|\frac{1+1}{1}\right|\right)\right]$ 
= $\left[-\frac{1}{3}+\log \left|\frac{4}{3}\right|+\left(1-\log \left|\frac{2}{1}\right|\right)\right]$ 
= $\left[-\frac{1}{3}+1+\log \left|\frac{4}{3} \times \frac{1}{2}\right|\right]$ 
$\Rightarrow \mathrm{I}=\left[\frac{2}{3}+\log | \frac{2}{3}\right] |$ 
$\text{L.H.S = R.H.S}$
Hence proved.

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