Question

lf $\int \frac{dx}{x-x^3}=$ Alog $|\frac{x^2}{1-x^2}|+c$ then $A=$

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{1}{2}$

$\int \frac{dx}{x-x^3}=\int \frac{dx}{x(1-x^2)}=\int \frac{dx}{x(1-x)(1+x)}$

$\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x};A=1;B=\frac{1}{2};C=\frac{-1}{2}$

$=\frac{1}{x}+=\frac{{(}^1{/}_2)}{1-x}+\frac{{(}^{-1}{/}_2)}{1+x}$

$\int [1x+\frac{{}^1{/}_2}{1-x}+\frac{{}^{-1}{/}_2}{1+x}]dx=logx-\frac{1}{2}log(1-x)$ $-\frac{1}{2}log(1+x)+c$

$logx-\left[log\right(1-x{}^2{)}^{{}^1{/}_2}]+c$

$=log\left|\frac{x}{(1-x^2{)}^{{}^1{/}_2}}\right|+c$

$=\frac{1}{2}log\left|\frac{x^2}{1-x^2}\right|+c$

$A=\frac{1}{2}$