Answer
\[\frac{1}{2}\ln |1+x^2|+\tan^{-1} x+C\]
Work Step by Step
Let \[I=\int\frac{1+x}{1+x^2}\:dx\]
\[I=\int\frac{\frac{1}{2}(2x)+1}{1+x^2}\:dx\]
\[I=\frac{1}{2}\int\frac{2x}{1+x^2}\:dx+\int\frac{dx}{1+x^2}\;\;\;...(1)\]
Let \[I_1=\frac{1}{2}\int\frac{2x}{1+x^2}\:dx\]
Put $t=1+x^2
\;\;\Rightarrow \;\;dt=2x\:dx$
\[I_1=\frac{1}{2}\int\frac{dt}{t}\]
\[I_1=\frac{1}{2}\ln |t|\]
\[I_1=\frac{1}{2}\ln |1+x^2|\;\;\;...(2)\]
Let \[I_2=\int\frac{1}{1+x^2}\:dx\]
\[I_2=\tan^{-1}x\;\;\;...(3)\]
Using (2) and (3) in (1)
\[I=\frac{1}{2}\ln |1+x^2|+\tan^{-1} x+C\]
Where $C$ is constant of integration
Hence, \[I=\frac{1}{2}\ln |1+x^2|+\tan^{-1} x+C\]
