Question:
$\int_{0}^{1} \frac{x}{x^{2}+1} d x$
Solution:
$\int_{0}^{1} \frac{x}{x^{2}+1} d x$
Let $x^{2}+1=t \Rightarrow 2 x d x=d t$
When $x=0, t=1$ and when $x=1, t=2$
$\therefore \int_{0}^{1} \frac{x}{x^{2}+1} d x=\frac{1}{2} \int_{1}^{2} \frac{d t}{t}$
$=\frac{1}{2}[\log |t|]_{1}^{2}$
$=\frac{1}{2}[\log 2-\log 1]$
$=\frac{1}{2} \log 2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.