Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Prove that $\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
✓
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)
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$
$\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}$
$\Rightarrow \int \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\int-\frac{1}{x} d x+\int \frac{1}{\left(x^{2}\right)} d x+\int \frac{1}{(x+1)} d x$
$\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\left[-\log |x|-x^{-1}+\log |x+1|\right]_{1}^{3}$
$\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x$ = $\left[-\frac{1}{x}+\log \left|\frac{x+1}{x}\right|\right]_{1}^{3}$
= $\left[-\frac{1}{3}+\log \left|\frac{3+1}{3}\right|-\left(-\frac{1}{1}+\log \left|\frac{1+1}{1}\right|\right)\right]$
= $\left[-\frac{1}{3}+\log \left|\frac{4}{3}\right|+\left(1-\log \left|\frac{2}{1}\right|\right)\right]$
= $\left[-\frac{1}{3}+1+\log \left|\frac{4}{3} \times \frac{1}{2}\right|\right]$
$\Rightarrow \mathrm{I}=\left[\frac{2}{3}+\log | \frac{2}{3}\right] |$
L.H.S = R.H.S
Hence proved.
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