# How do you integrate int x/sqrt(2+3x) by parts?

Nov 17, 2016

$\int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{2}{27} \sqrt{2 + 3 x} \left(3 x - 4\right) + C$

#### Explanation:

You would not integrate this integral By Parts, instead you would use a substitution ;

Let $u = 2 + 3 x \implies x = \frac{1}{3} \left(u - 2\right)$
And, $\frac{\mathrm{du}}{\mathrm{dx}} = 3 \implies \frac{1}{3} \int \ldots \mathrm{du} = \int \ldots \mathrm{dx}$

Applying this substitution to the problem gives us:

$\int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \int \frac{\frac{1}{3} \left(u - 2\right)}{\sqrt{u}} \left(\frac{1}{3}\right) \mathrm{du}$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{1}{9} \int \left(u - 2\right) {u}^{- \frac{1}{2}} \mathrm{du}$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{1}{9} \int {u}^{\frac{1}{2}} - 2 {u}^{- \frac{1}{2}} \mathrm{du}$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{1}{9} \left\{{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - \frac{2 {u}^{\frac{1}{2}}}{\frac{1}{2}}\right\} + C$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{1}{9} {u}^{\frac{1}{2}} \left\{\frac{2}{3} u - 4\right\} + C$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{1}{9} {u}^{\frac{1}{2}} \left\{\frac{2}{3} \left(u - 6\right)\right\} + C$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{2}{27} {u}^{\frac{1}{2}} \left(u - 6\right) + C$

Substituting for $u$ gives is;

$\int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{2}{27} \sqrt{2 + 3 x} \left(2 + 3 x - 6\right) + C$
$\therefore \int \frac{x}{\sqrt{2 + 3 x}} \mathrm{dx} = \frac{2}{27} \sqrt{2 + 3 x} \left(3 x - 4\right) + C$

Nov 17, 2016

#### Explanation:

$I = \int x {\left(2 + 3 x\right)}^{- \frac{1}{2}} \mathrm{dx}$

Let $u = x$ and $\mathrm{dv} = {\left(2 + 3 x\right)}^{- \frac{1}{2}} \mathrm{dx}$

This make $\mathrm{du} = \mathrm{dx}$ and

$v = \int {\left(2 + 3 x\right)}^{- \frac{1}{2}} \mathrm{dx} = \frac{2}{3} {\left(2 + 3 x\right)}^{\frac{1}{2}}$ (by substitution)

$I = u v - \int v \mathrm{du}$

$= \frac{2}{9} x {\left(2 + 3 x\right)}^{\frac{1}{2}} - \frac{2}{3} \int {\left(2 + 3 x\right)}^{\frac{1}{2}} \mathrm{dx}$

$= \frac{2}{3} x {\left(2 + 3 x\right)}^{\frac{1}{2}} - \frac{2}{3} \left[\frac{2}{9} {\left(2 + 3 x\right)}^{\frac{3}{2}}\right] + C$ (by substitution)

Clean up the algebra as desired.