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

Jan 6, 2017

The answer is $\frac{1}{20} {\left(2 x - 1\right)}^{\frac{5}{2}} + \frac{1}{6} {\left(2 x - 1\right)}^{\frac{3}{2}} - \frac{3}{4} {\left(2 x - 1\right)}^{\frac{1}{2}} + C$

#### Explanation:

Let $u = 2 x - 1$. Then $\mathrm{du} = 2 \mathrm{dx}$ and $\mathrm{dx} = \frac{1}{2} \mathrm{du}$.

$\int \frac{{x}^{2} - 1}{\sqrt{u}} \cdot \frac{1}{2} \mathrm{du}$

We want to get rid of the x's. $x = \frac{u + 1}{2}$, so ${x}^{2} = {\left(\frac{u + 1}{2}\right)}^{2}$

$\int \frac{{\left(\frac{u + 1}{2}\right)}^{2} - 1}{\sqrt{u}} \cdot \frac{1}{2} \mathrm{du}$

Expand:

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

$\frac{1}{2} \int \frac{{u}^{2} + 2 u + 1 - 4}{4 \sqrt{u}} \mathrm{du}$

$\frac{1}{8} \int \frac{{u}^{2} + 2 u - 3}{\sqrt{u}}$

$\frac{1}{8} \int \left({u}^{2} / \sqrt{u} + \frac{2 u}{\sqrt{u}} - \frac{3}{\sqrt{u}}\right) \mathrm{du}$

$\frac{1}{8} \int \left({u}^{\frac{3}{2}} + 2 {u}^{\frac{1}{2}} - 3 {u}^{- \frac{1}{2}}\right) \mathrm{du}$

$\frac{1}{8} \left(\frac{2}{5} {u}^{\frac{5}{2}} + \frac{4}{3} {u}^{\frac{3}{2}} - 6 {u}^{\frac{1}{2}}\right)$

$\frac{1}{20} {u}^{\frac{5}{2}} + \frac{1}{6} {u}^{\frac{3}{2}} - \frac{3}{4} {u}^{\frac{1}{2}}$

$\frac{1}{20} {\left(2 x - 1\right)}^{\frac{5}{2}} + \frac{1}{6} {\left(2 x - 1\right)}^{\frac{3}{2}} - \frac{3}{4} {\left(2 x - 1\right)}^{\frac{1}{2}} + C$

Hopefully this helps!

Nov 15, 2017

$\int \frac{{x}^{2} - 1}{\sqrt{2 x - 1}} \cdot \mathrm{dx}$

=$\frac{1}{4} \cdot \int \frac{4 {x}^{2} - 4}{\sqrt{2 x - 1}} \cdot \mathrm{dx}$

=$\frac{1}{4} \cdot \int \frac{{\left(2 x\right)}^{2} - 4}{\sqrt{2 x - 1}} \cdot \mathrm{dx}$

After using $u = \sqrt{2 x - 1}$, $2 x = {u}^{2} + 1$, $2 \mathrm{dx} = 2 u \cdot \mathrm{du}$ or $\mathrm{dx} = u \cdot \mathrm{du}$ transforms, this integral became

$\frac{1}{4} \cdot \int \frac{{\left(2 x\right)}^{2} - 4}{\sqrt{2 x - 1}} \cdot \mathrm{dx}$

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

=$\frac{1}{4} \cdot \int \left({u}^{4} + 2 {u}^{2} - 3\right) \cdot \mathrm{du}$

=$\frac{1}{4} \cdot \left({u}^{5} / 5 + \frac{2 {u}^{3}}{3} - 3 u\right) + C$

=${u}^{5} / 20 + {u}^{3} / 6 - \frac{3 u}{4} + C$

=$\frac{1}{20} \cdot {\left(2 x - 1\right)}^{\frac{5}{2}} + \frac{1}{6} \cdot {\left(2 x - 1\right)}^{\frac{3}{2}} - \frac{3}{4} \cdot \sqrt{2 x - 1} + C$