Question
Integrate the function $\frac{1}{x-x^{3}}$
✓
Answer
Given $\frac{1}{x-x^{3}}$
Let $I=\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1+x)(1-x)}$
Using partial fraction:
Let $\frac{1}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1-x} ...(i)$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A(1+x)(1-x)+B(x)(1-x)+C(x)(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow 1 = A - Ax^2 + Bx - Bx^2 + Cx + Cx^2$
$\Rightarrow 1 = A + (B + C)x + (-A - B + C)x^2$
Equating the coefficients of $x, x^2$ and constant value. We get:
$ A = 1, $
$B + C = 0 $
$\Rightarrow B = -C$
$-A - B + C = 0$
$\Rightarrow -1 - (- C) +C = 0$
$\Rightarrow 2C = 1 $
$\Rightarrow C = \frac{1}{2}$
So,$ B = -\frac{1}{2}$
Put these values in equation $(i)$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{1}{x}+\frac{-\left(\frac{1}{2}\right)}{1+x}+\frac{\left(\frac{1}{2}\right)}{1-x}$
$\Rightarrow \int \frac{1}{x(1+x)(1-x)} dx$ = $\int \frac{1}{x} d x-\frac{1}{2} \int \frac{1}{1+x} d x+\frac{1}{2} \int \frac{1}{1-x} d$
$= \log |\mathrm{x}|-\frac{1}{2} \log |1+\mathrm{x}|+\frac{1}{2} \log |1-\mathrm{x}|$
$= \log |x|-\log \left|(1+x)^{\frac{1}{2}}\right|+\log \left|(1-x)^{\frac{1}{2}}\right|$
$= \log \left|\frac{x}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{1}{2}}}\right|+C$
$= \log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{(1+x)(1-x)^{\frac{1}{2}}}\right|+C$
$= \log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{\left(1-x^{2}\right)^{\frac{1}{2}}}\right|+C$
$= \log \left|\left(\frac{x^{2}}{1-x^{2}}\right)^{\frac{1}{2}}\right|+C$
$\Rightarrow \mathrm{I}=\frac{1}{2} \log \left|\frac{\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right|+\mathrm{C}$
Let $I=\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1+x)(1-x)}$
Using partial fraction:
Let $\frac{1}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1-x} ...(i)$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A(1+x)(1-x)+B(x)(1-x)+C(x)(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow 1 = A - Ax^2 + Bx - Bx^2 + Cx + Cx^2$
$\Rightarrow 1 = A + (B + C)x + (-A - B + C)x^2$
Equating the coefficients of $x, x^2$ and constant value. We get:
$ A = 1, $
$B + C = 0 $
$\Rightarrow B = -C$
$-A - B + C = 0$
$\Rightarrow -1 - (- C) +C = 0$
$\Rightarrow 2C = 1 $
$\Rightarrow C = \frac{1}{2}$
So,$ B = -\frac{1}{2}$
Put these values in equation $(i)$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{1}{x}+\frac{-\left(\frac{1}{2}\right)}{1+x}+\frac{\left(\frac{1}{2}\right)}{1-x}$
$\Rightarrow \int \frac{1}{x(1+x)(1-x)} dx$ = $\int \frac{1}{x} d x-\frac{1}{2} \int \frac{1}{1+x} d x+\frac{1}{2} \int \frac{1}{1-x} d$
$= \log |\mathrm{x}|-\frac{1}{2} \log |1+\mathrm{x}|+\frac{1}{2} \log |1-\mathrm{x}|$
$= \log |x|-\log \left|(1+x)^{\frac{1}{2}}\right|+\log \left|(1-x)^{\frac{1}{2}}\right|$
$= \log \left|\frac{x}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{1}{2}}}\right|+C$
$= \log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{(1+x)(1-x)^{\frac{1}{2}}}\right|+C$
$= \log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{\left(1-x^{2}\right)^{\frac{1}{2}}}\right|+C$
$= \log \left|\left(\frac{x^{2}}{1-x^{2}}\right)^{\frac{1}{2}}\right|+C$
$\Rightarrow \mathrm{I}=\frac{1}{2} \log \left|\frac{\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right|+\mathrm{C}$
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