# Thread: Integration using U sub

1. ## Integration using U sub

I know its U substitution with u= 4+3s^2 and du=6xdx but I dont know how to make 13 s drop out?
Thanks to anyone who can help
Amanda

2. $du=6s \, ds \Longrightarrow \frac{13}{6}du=13s \, ds.$

3. Originally Posted by Casas4
I know its U substitution with u= 4+3s^2 and du=6xdx ...
Which means $dx = \dfrac{du}{6x}$. Substituting $\dfrac{du}{6x}$ for $dx$ and $u$ for $4+3s^2$ gives: $\int\left(\dfrac{13s}{u}\right)\left(\dfrac{du}{6s }\right)\;{du} = \int\left(\dfrac{13s}{6s}\right)\left(\dfrac{1}{u} \right)\;{du}$. The $s$ in the numerator and the $s$ in the denominator cancel, so it becomes: $\frac{13}{6}\int\frac{1}{u}du$.