# \intx(2x+5)^8dx?

## so far I've worked out the u-sub to be $u = 2 x + 5$, so $\mathrm{du} = 2 \mathrm{dx} \setminus \Rightarrow \mathrm{dx} = \frac{\mathrm{du}}{2}$

Sep 16, 2017

$\int x {\left(2 x + 5\right)}^{8} \mathrm{dx} = \frac{1}{360} \left(18 x - 5\right) {\left(2 x + 5\right)}^{9}$

#### Explanation:

Your substitution is good, all you need to do is complete it and then integrate term by term.

$\int x {\left(2 x + 5\right)}^{8} \mathrm{dx} = \frac{1}{2} \int \frac{1}{2} \left(u - 5\right) {u}^{8} \mathrm{du} = \frac{1}{4} \int {u}^{9} - 5 {u}^{8} \mathrm{du} = \frac{1}{4} \left(\frac{1}{10} {u}^{10} - \frac{5}{9} {u}^{9}\right) + \text{c" = 1/360(9(2x+5)^10-50(2x+5)^9)) +"c} = \frac{1}{360} \left(18 x - 5\right) {\left(2 x + 5\right)}^{9}$

Sep 16, 2017

See below.

#### Explanation:

$u = 2 x + 8$, so $\mathrm{du} = 2 \mathrm{dx} \setminus \Rightarrow \mathrm{dx} = \frac{\mathrm{du}}{2}$

$\setminus \int x {\left(2 x + 8\right)}^{8} \mathrm{dx} \equiv \int \frac{u - 8}{2} {u}^{8} \frac{\mathrm{du}}{2} = \frac{1}{4} \int {u}^{9} \mathrm{du} - \frac{8}{4} \int {u}^{8} \mathrm{du} = \frac{1}{40} {u}^{10} - \frac{8}{36} {u}^{9}$

and now

$\setminus \int x {\left(2 x + 8\right)}^{8} \mathrm{dx} = \frac{1}{40} {\left(2 x + 8\right)}^{10} - \frac{8}{36} {\left(2 x + 8\right)}^{9} + C$

Sep 17, 2017

$\frac{1}{360} {\left(2 x + 5\right)}^{9} \left(18 x - 5\right) + C .$

#### Explanation:

It is confirmed from the Questioner that, the Problem is,

$\int x {\left(2 x + 5\right)}^{8} \mathrm{dx} .$

Let us solve the Problem without using the Method of Substitution.

Note that,

$\int {\left(a x + b\right)}^{n} \mathrm{dx} = \frac{1}{a} \cdot {\left(a x + b\right)}^{n + 1} / \left(n + 1\right) + c , a \ne 0 , n \ne - 1. . . \left(\star\right)$

Suppose that, $I = \int x {\left(2 x + 5\right)}^{8} \mathrm{dx} ,$

$\therefore I = \frac{1}{2} \int \left(2 x\right) {\left(2 x + 5\right)}^{8} \mathrm{dx} ,$

$= \frac{1}{2} \int \left\{\left(2 x + 5\right) - 5\right\} {\left(2 x + 5\right)}^{8} \mathrm{dx} ,$

$= \frac{1}{2} \int \left\{\left(2 x + 5\right) {\left(2 x + 5\right)}^{8} - 5 {\left(2 x + 5\right)}^{8}\right\} \mathrm{dx} ,$

$= \frac{1}{2} \int \left\{{\left(2 x + 5\right)}^{9} - 5 {\left(2 x + 5\right)}^{8}\right\} \mathrm{dx} ,$

$= \frac{1}{2} \int {\left(2 x + 5\right)}^{9} \mathrm{dx} - \frac{5}{2} \int {\left(2 x + 5\right)}^{8} \mathrm{dx} .$

Using $\left(\star\right) ,$ now, we get,

$I = \frac{1}{2} \cdot \frac{1}{2} {\left(2 x + 5\right)}^{9 + 1} / \left(9 + 1\right) - \frac{5}{2} \cdot \frac{1}{2} {\left(2 x + 5\right)}^{8 + 1} / \left(8 + 1\right) ,$

$= \frac{1}{40} {\left(2 x + 5\right)}^{10} - \frac{5}{36} {\left(2 x + 5\right)}^{9.}$

$\Rightarrow I = \frac{1}{360} {\left(2 x + 5\right)}^{9} \left(18 x - 5\right) + C .$

Enjoy Maths.!