# Integrate the following using infinite \bb\text(series) ?

## (a) $\setminus \int \frac{\setminus \arctan \left(4 x\right)}{{x}^{-} 7} \mathrm{dx}$

May 10, 2018

$\int \arctan \frac{4 x}{x} ^ \left(- 7\right) \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {2}^{4 n + 2} / \left(\left(2 n + 1\right) \left(2 n + 9\right)\right) {x}^{2 n + 9} + C$

#### Explanation:

Start from:

$\arctan \left(4 x\right) = {\int}_{0}^{4 x} \frac{\mathrm{dt}}{1 + {t}^{2}}$

Expand now the integrand using the geometric series:

$\arctan \left(4 x\right) = {\int}_{0}^{4 x} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{2 n} \mathrm{dt}$

for $\left\mid 4 x \right\mid < 1$ the series is absolutely convergent and we can integrate term by term:

$\arctan \left(4 x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\int}_{0}^{4 x} {t}^{2 n} \mathrm{dt}$

$\arctan \left(4 x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left(4 x\right)}^{2 n + 1} / \left(2 n + 1\right)$

$\arctan \left(4 x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {2}^{4 n + 2} / \left(2 n + 1\right) {x}^{2 n + 1}$

Dividing by ${x}^{- 7}$ means multiplying by ${x}^{7}$ and we can do it again term by term:

$\arctan \frac{4 x}{x} ^ \left(- 7\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {2}^{4 n + 2} / \left(2 n + 1\right) {x}^{2 n + 8}$

and integrating again term by term:

$\int \arctan \frac{4 x}{x} ^ \left(- 7\right) \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {2}^{4 n + 2} / \left(2 n + 1\right) \int {x}^{2 n + 8} \mathrm{dx}$

$\int \arctan \frac{4 x}{x} ^ \left(- 7\right) \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {2}^{4 n + 2} / \left(\left(2 n + 1\right) \left(2 n + 9\right)\right) {x}^{2 n + 9} + C$