Question

Integrate $\frac{1}{x^3(x^2-1)}$

• A. $\frac{1}{3}\log \left|\frac{x^3}{x^3-1}\right|+C$
• B. $\frac{1}{3}\log \left|\frac{1-x^3}{x^3}\right|+C$
• C. $\log \left|\frac{x^3}{x^3-1}\right|+C$
• D. $\frac{1}{3}\log \left|\frac{x^3-1}{x^3}\right|+C$

Answer 1

Let $l=\int \frac{1}{x^3(x^3-1)}dx$
$=\int \frac{1}{x^3-1}dx-\int \frac{1}{x^3}dx$
$=\int \frac{1}{(x-1)(x^2+x+1)}dx+\frac{1}{2x^2}$
$=\int \left[\frac{1}{3(x-1)}+\frac{\frac{-1}{3}x-\frac{2}{3}}{x^2+x+1}\right]dx+\frac{1}{2x^2}$
$=\int \frac{1}{3(x-1)}dx-\frac{1}{3}\int \frac{x+2}{x^2+x+1}dx+\frac{1}{a{x}^2}$
$=\frac{1}{3}\log (x-1)+\frac{1}{2x^2}-\frac{1}{3}\int \frac{\frac{1}{2}(2x+1)+\frac{3}{2}}{x^2+x+1}dx$
$=\frac{1}{3}\log (x-1)+\frac{1}{2x^2}-\frac{1}{6}\int \frac{2x+1}{x^2+x+1}dx$ $-\frac{1}{2}\int \frac{dx}{x^2+x+1}$
$=\frac{1}{3}\log (x-1)+\frac{1}{2x^2}-\frac{1}{6}\log (x^2+x+1)$ $-\frac{1}{2}\int \frac{dx}{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=\frac{1}{3}\log (x-1)+\frac{1}{2x^2}-\frac{1}{6}\log (x^2+x+1)$ $-\frac{1}{2}.\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$
$=\frac{1}{2}\log (x-1)+\frac{1}{2x^2}-\frac{1}{6}\log (x^2+x+1)$ $-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$