# Thread: Evaluating Indefinate Integerals

1. ## Evaluating Indefinate Integerals

#37, I know to make u = 1 + x but I am lost after that..

https://imgur.com/a/VZXox

2. ## Re: Evaluating Indefinate Integerals

Originally Posted by MrJank
#37, I know to make u = 1 + x but I am lost after that..

https://imgur.com/a/VZXox
If $u=1+x$ then $\dfrac{{xdx}}{{\sqrt {1 + x} }} = \dfrac{{u - 1}}{{\sqrt u }}du = \left( {\sqrt u - \dfrac{1}{{\sqrt u }}} \right)du$

3. ## Re: Evaluating Indefinate Integerals

Originally Posted by MrJank
#37, I know to make u = 1 + x

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$

4. ## Re: Evaluating Indefinate Integerals

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.

5. ## Re: Evaluating Indefinate Integerals

Originally Posted by HallsofIvy
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.
Or at least differing only by a constant :P