Question

$\underset{0}{\overset{1}{\int }}\frac{2x}{1+x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^1\frac{2x}{1+x^2}dx \\ \mathrm{Putting}1+x^2=t \\ \Rightarrow 2xdx=dt \\ \mathrm{When}x\to 0;t\to 1 \\ \mathrm{And}x\to 1;t\to 2 \\ \therefore I={\int }_1^2\frac{dt}{t} \\ ={\left[{\log }_e\left|t\right|\right]}_1^2 \\ ={\log }_{e}2-{\log }_{e}1 \\ ={\log }_{e}2-0 \\ ={\log }_{e}2 \end{gathered}$$ 4430. →