∫ 1 + x 1 + x 2 d x = ∫ ( 1 1 + x 2 + x 1 + x 2 ) d x = ∫ 1 1 + x 2 d x + ∫ x 1 + x 2 d x = t a n − 1 ( x ) + 1 2 ∫ 1 u d u = t a n − 1 ( x ) + ln | 1 + x | + c {\displaystyle {\begin{aligned}\int {\frac {1+x}{1+x^{2}}}dx&=\int ({\frac {1}{1+x^{2}}}+{\frac {x}{1+x^{2}}})dx\\[2ex]&=\int {\frac {1}{1+x^{2}}}dx+\int {\frac {x}{1+x^{2}}}dx\\[2ex]&=tan^{-1}(x)+{\frac {1}{2}}\int {\frac {1}{u}}du\\[2ex]&=tan^{-1}(x)+\ln |{1+x}|+c\end{aligned}}}
![{\displaystyle {\begin{aligned}\int {\frac {1+x}{1+x^{2}}}dx&=\int ({\frac {1}{1+x^{2}}}+{\frac {x}{1+x^{2}}})dx\\[2ex]&=\int {\frac {1}{1+x^{2}}}dx+\int {\frac {x}{1+x^{2}}}dx\\[2ex]&=tan^{-1}(x)+{\frac {1}{2}}\int {\frac {1}{u}}du\\[2ex]&=tan^{-1}(x)+\ln |{1+x}|+c\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cae5cc284ce45518a7c5eda74b1f2146b537c547)