Integrals — Class 12 Maths Solution

Let $I = \int\limits_0^1 {\cfrac{x}{{{x^2} + 1}}} dx$
Put ${x^2} + 1 = t$ $\Rightarrow$ $2xdx = dt$

When $x = 0,t = 1$ and when $x = 1,t = 2$

$\therefore$ $I = \cfrac{1}{2}\int\limits_1^2 {\cfrac{{dt}}{t} = \left[ {\cfrac{1}{2}\log t} \right]_1^2 = \cfrac{1}{2}\left[ {\log 2 - \log 1} \right]} = \cfrac{1}{2}\log 2$