Integrate the function 1 x-x^ 3

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)
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A(1+x)(1-x)+B(x)(1-x)+C(x)(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1-x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}$
⇒ 1 = A - Ax² + Bx - Bx² + Cx + Cx²
⇒ 1 = A + (B + C)x + (-A - B + C)x²
Equating the coefficients of x, x² and constant value. We get:
A = 1,
B + C = 0 $\Rightarrow$ B = -C
-A - B + C = 0
⇒ -1 - (- C) +C = 0
⇒ 2C = 1 ⇒ C = $\frac{1}{2}$
So, B = $-\frac{1}{2}$
Put these values in equation (i)
$\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{1}{x}+\frac{-\left(\frac{1}{2}\right)}{1+x}+\frac{\left(\frac{1}{2}\right)}{1-x}$
$\Rightarrow \int \frac{1}{x(1+x)(1-x)} dx$ = $\int \frac{1}{x} d x-\frac{1}{2} \int \frac{1}{1+x} d x+\frac{1}{2} \int \frac{1}{1-x} d$
= $\log |\mathrm{x}|-\frac{1}{2} \log |1+\mathrm{x}|+\frac{1}{2} \log |1-\mathrm{x}|$
= $\log |x|-\log \left|(1+x)^{\frac{1}{2}}\right|+\log \left|(1-x)^{\frac{1}{2}}\right|$
= $\log \left|\frac{x}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{1}{2}}}\right|+C$
= $\log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{(1+x)(1-x)^{\frac{1}{2}}}\right|+C$
= $\log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{\left(1-x^{2}\right)^{\frac{1}{2}}}\right|+C$
= $\log \left|\left(\frac{x^{2}}{1-x^{2}}\right)^{\frac{1}{2}}\right|+C$
$\Rightarrow \mathrm{I}=\frac{1}{2} \log \left|\frac{\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right|+\mathrm{C}$