# How do you integrate ((x^2-1)/sqrt(2x-1) )dx?

May 28, 2018

$I = \frac{1}{60} \left[3 {\left(\sqrt{2 x - 1}\right)}^{5} + 10 {\left(\sqrt{2 x - 1}\right)}^{3} - 45 \sqrt{2 x - 1}\right] + c$

#### Explanation:

Here,

$I = \int \left(\frac{{x}^{2} - 1}{\sqrt{2 x - 1}}\right) \mathrm{dx}$

Subst, $\sqrt{2 x - 1} = u \implies 2 x - 1 = {u}^{2} \implies 2 x = {u}^{2} + 1$

$\implies x = \frac{1}{2} \left({u}^{2} + 1\right) \implies \mathrm{dx} = \frac{1}{2} \left(2 u\right) \mathrm{du} = u \mathrm{du}$

So,

$I = \int \frac{\frac{1}{4} {\left({u}^{2} + 1\right)}^{2} - 1}{u} \times u \mathrm{du}$

$I = \int \left[\frac{1}{4} \left({u}^{4} + 2 {u}^{2} + 1\right) - 1\right] \mathrm{du}$

$= \frac{1}{4} \int \left[{u}^{4} + 2 {u}^{2} + 1 - 4\right] \mathrm{du}$

$= \frac{1}{4} \int \left[{u}^{4} + 2 {u}^{2} - 3\right] \mathrm{du}$

$= \frac{1}{4} \left[{u}^{5} / 5 + 2 {u}^{3} / 3 - 3 u\right] + c$

$= \frac{1}{4} \times \frac{1}{15} \left[3 {u}^{5} + 10 {u}^{3} - 45 u\right] + c$

$= \frac{1}{60} \left[3 {\left(u\right)}^{5} + 10 {\left(u\right)}^{3} - 45 u\right] + c$

Subst, back , $u = \sqrt{2 x - 1}$

$I = \frac{1}{60} \left[3 {\left(\sqrt{2 x - 1}\right)}^{5} + 10 {\left(\sqrt{2 x - 1}\right)}^{3} - 45 \sqrt{2 x - 1}\right] + c$