Mathematics

Solve:
$$\int {\dfrac{1}{x-x^3}}dx$$


SOLUTION
$$ I = \int \dfrac{1}{x(1-x^{2})}dx = \int \dfrac{1}{x(1-x)(1+x)}dx$$

using parctial fraction, we have
$$ I = \int \dfrac{A}{x}+\dfrac{B}{1-x} +\dfrac{C}{1+x}$$

on comparing numerator, we have

In $$ A(1-x)(1+x)+Bx(1+x)+C(1-x)x = 1 $$

put $$x =1, B = 1/2 \ put \ x = 0, A = 1$$

put $$x = -1, C = -1/2$$

$$ \Rightarrow I = \int (\dfrac{1}{x}+\dfrac{1}{2(1-x)}-\dfrac{1}{2(1+x)})dx$$

$$ = lnx -\dfrac{1}{2}ln(1-x)+\dfrac{1}{2}ln(1+x)$$

$$ = ln(\dfrac{x\sqrt{1+x}}{\sqrt{1-x}})$$ 
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Subjective Medium Published on 17th 09, 2020
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