Can anyone give me any advice on solving this integral?
((x^2-1))/((square root(2x-1))
Would I use u=x^2-1 then du=2x ?
It's often easier to pick your sub as what's in the square root as this gives the nicer $\displaystyle \sqrt{u}$ when it comes to integrating.
$\displaystyle \dfrac{x^2-1}{\sqrt{2x-1}} = \dfrac{x^2}{\sqrt{2x-1}} - \dfrac{1}{\sqrt{2x-1}}$
For the first term sub $\displaystyle u = 2x-1 \Leftrightarrow du = 2 dx$ and also $\displaystyle x = \dfrac{1}{2}(u+1) \Leftrightarrow x^2= \dfrac{1}{4}(u+1)^2 = \dfrac{1}{4}(u^2+2u+1)$
This gives a first term of $\displaystyle \dfrac{1}{4} \int \dfrac{u^2+2u+1}{\sqrt{u}}\ du = \dfrac{1}{4}\left( \int \dfrac{u^2}{\sqrt{u}}\ du + \int \dfrac{2u}{\sqrt{u}}\ du + \int \dfrac{1}{\sqrt{u}}\ du\right)$
When that's integrated we get $\displaystyle \dfrac{1}{4} \left(\dfrac{2}{5}u^{5/2} + \dfrac{4}{3}u^{3/2} + 2u^{1/2} + C'\right) = \dfrac{1}{10}u^{5/2} + \dfrac{1}{3}u^{3/2} + \dfrac{1}{2}\sqrt{u} + C_1$
where $\displaystyle C_1 = \dfrac{1}{4}C'$ which is the constant of integration.
Can you integrate the second term using the same u-sub
Hello, homeylova223!
$\displaystyle \int \frac{x^2-1}{\sqrt{2x-1}}\,dx$
When the radicand is linear, I let $\displaystyle u$ equal the entire radical.
$\displaystyle \text{Let: }\,u \,=\,\sqrt{2x-1} \quad\Rightarrow\quad x \,=\,\frac{u^2+1}{2} \quad\Rightarrow\quad dx \,=\,u\,du$
. . $\displaystyle x^2-1 \:=\:\left(\frac{u^2+1}{2}\right)^2 - 1 \:=\:\frac{u^4 + 2u^2 - 3}{4}$
Substitute: .$\displaystyle \int\frac{\frac{u^4+2u^2 - 3}{4}}{u}\,u\,du \;=\;\tfrac{1}{4}\int(u^4 + 2u^2 - 3)\,du $
. . $\displaystyle =\;\tfrac{1}{4}\left(\tfrac{1}{5}u^5 + \tfrac{2}{3}u^3 - 3u\right) + C \;=\;\tfrac{u}{60}\left(3u^4 + 10u^2 - 45) + C $
Back-substitute: .$\displaystyle \frac{\sqrt{2x-1}}{60}\bigg[3(2x-1)^2 + 10(2x-1) - 45\bigg] + C $
. . . . . . . . . . . $\displaystyle =\;\frac{1}{60}\sqrt{2x-1}\,(12x^2 + 8x - 52) + C$
. . . . . . . . . . . $\displaystyle =\;\frac{1}{15}\sqrt{2x-1}\,(3x^2 + 2x-13) + C $