∫ a + b x 2 3 a x + b x 3 d x {\displaystyle \int {\frac {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx}
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} u &= \3ax+bx^3 \\[2ex] du &= \frac{1}{x}dx \\[2ex] \end{align} }
∫ 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}\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)