#37, I know to make u = 1 + x but I am lost after that..
https://imgur.com/a/VZXox
#37, I know to make u = 1 + x but I am lost after that..
https://imgur.com/a/VZXox
No, you are not required to make u = 1 + x. Be open to other expressions.
$\displaystyle \displaystyle\int\dfrac{x}{\sqrt{1 + x}}dx$
For example,
let $\displaystyle \ \ u \ = \ \sqrt{1 + x}$.
Then $\displaystyle \ \ u^2 \ = \ 1 + x$.
$\displaystyle x \ = \ u^2 - 1 $
$\displaystyle dx \ = \ 2u \ du $
Substitute:
$\displaystyle \displaystyle\int \bigg(\dfrac{u^2 - 1}{u}\bigg)2u \ du$
$\displaystyle \displaystyle\int\(2u^2 \ - \ 2)du$
It is an interesting exercise to do this problem with the substitution u= x+ 1, so integrating $\displaystyle \int u^{1/2}- u^{-1/2} du$ as Plato gives, then doing the problem with $\displaystyle u= \sqrt{x+ 1}$, so integrating $\displaystyle \int 2u^2- 2 du$, as greg1313 suggests, and showing that the results are the same.