Question

Evaluate the integral
${\int }_0^1\frac{x}{1+x^2}dx$

• A. $\log 2$
• B. $\frac{1}{2}\log 2$
• C. $2$
• D. $\log 4$

Answer 1

The correct option is B $\frac{1}{2}\log 2$

${\int }_0^1\frac{x}{a+x^2}dx$

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

$={\int }_0^1\frac{x}{1+x^2}dx={\left[\frac{1}{2}\log \right(1+x^2)+c]}_0^1$

$=\left(\frac{1}{2}\log \right(2)+c)-\left(\frac{1}{2}\log \right(1)+c)$

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

${\int }_0^1\frac{x}{1+x^2}dx=\frac{1}{2}\log 2$