∫ sin ( ln ( x ) ) x d x {\displaystyle \int {\frac {\sin {(\ln {(x))}}}{x}}dx}
u = ln ( x ) d u = 1 x d x {\displaystyle {\begin{aligned}u&=\ln(x)\\[2ex]du&={\frac {1}{x}}dx\\[2ex]\end{aligned}}}
∫ x 3 ( 2 + x 4 ) 5 d x = ∫ ( x 3 d x ) ( 2 + x 4 ) = ∫ ( 1 4 d u ) ( u ) = 1 4 ∫ u d u = 1 4 [ u 1 + 1 1 + 1 ] + C = u 2 8 + C = ( 2 + x 4 ) 2 8 + C {\displaystyle {\begin{aligned}\int x^{3}(2+x^{4})^{5}dx&=\int (x^{3}dx)(2+x^{4})=\int \left({\frac {1}{4}}du\right)(u)={\frac {1}{4}}\int u\,du\\[2ex]&={\frac {1}{4}}\left[{\frac {u^{1+1}}{1+1}}\right]+C={\frac {u^{2}}{8}}+C\\[2ex]&={\frac {(2+x^{4})^{2}}{8}}+C\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=\ln(x)\\[2ex]du&={\frac {1}{x}}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/2dcd1cacb47fe0ded9f0cb8411297ff136d61465)
![{\displaystyle {\begin{aligned}\int x^{3}(2+x^{4})^{5}dx&=\int (x^{3}dx)(2+x^{4})=\int \left({\frac {1}{4}}du\right)(u)={\frac {1}{4}}\int u\,du\\[2ex]&={\frac {1}{4}}\left[{\frac {u^{1+1}}{1+1}}\right]+C={\frac {u^{2}}{8}}+C\\[2ex]&={\frac {(2+x^{4})^{2}}{8}}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/076298ca80fdae8f38ea211c551bb3425c836c4b)