∫ 0 7 4 + 3 x d x {\displaystyle \int _{0}^{7}{\sqrt {4+3x}}\,dx}
u = 4 + 3 x d u = 3 d x 1 3 d u = d x {\displaystyle {\begin{aligned}u&=4+3x\\[2ex]du&=3\,dx\\[2ex]{\frac {1}{3}}du&=dx\\[2ex]\end{aligned}}}
∫ sin ( ln ( x ) ) x d x = ∫ 1 x sin ( ln ( x ) ) d x = ∫ ( 1 x d x ) sin ( ln ( x ) ) = ∫ ( d u ) sin ( u ) = ∫ sin ( u ) d u = − cos ( u ) + C = − cos ( ln ( x ) ) + C {\displaystyle {\begin{aligned}\int {\frac {\sin {(\ln {(x))}}}{x}}dx&=\int {\frac {1}{x}}\sin(\ln {(x)})dx=\int \left({\frac {1}{x}}dx\right)\sin {(\ln {(x)})}\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=4+3x\\[2ex]du&=3\,dx\\[2ex]{\frac {1}{3}}du&=dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/75fbdf9385d28dbfa67df6bec234667b6be64498)
![{\displaystyle {\begin{aligned}\int {\frac {\sin {(\ln {(x))}}}{x}}dx&=\int {\frac {1}{x}}\sin(\ln {(x)})dx=\int \left({\frac {1}{x}}dx\right)\sin {(\ln {(x)})}\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/92ec083cbb58cbf42efe622bfc22fd4c76bb337f)