Question

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

• A. ${\tan }^{-1}x+\frac{1}{x}+c$
• B. ${\tan }^{-1}x-\frac{1}{x}+c$
• C. $\frac{{\tan }^{-1}x}{x}+c$
• D. $\frac{{\tan }^{-1}x}{x^2}+c$

Answer 1

The correct option is B ${\tan }^{-1}x-\frac{1}{x}+c$

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

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

$=\int [\frac{1+x^2}{x^2(1+x^2)}+\frac{x^2}{x^2(1+x^2)}]dx$

$=\int [\frac{1}{x^2}+\frac{1}{1+x^2}]dx$

$=\frac{-1}{x}+{\tan }^{-1}x+c$

$={\tan }^{-1}x-\frac{1}{x}+c$