Question

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

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

Answer 1

The correct option is B ${\tan }^{-1}x+\frac{1}{2}\log (1+x^2).$

$\int \frac{1+x}{1+x^2}dx=\int [\frac{1}{1+x^2}+\frac{x}{1+x^2}]dx$
$=\int [\frac{1}{1+x^2}+\frac{1}{2}.\frac{2x}{1+x^2}]dx$
Hence, option 'B' is correct.
$={\tan }^{-1}x+\frac{1}{2}\log (1+x^2)$