Solved: "How would one integrate the following?∫xn−2(1+x)ndx"

Anahi Solomon

Answered question

2022-03-30

How would one integrate the following?

\int \frac{x^{n - 2}}{{(1 + x)}^{n}} d x

armejantm925

Beginner2022-03-31Added 20 answers

\int \frac{x^{n - 2}}{{(1 + x)}^{n}} d x = \int {(1 + x)}^{- 2} {(\frac{x}{1 + x})}^{n - 2} d x

= \int y^{n - 2} d y

using the substitution

y = \frac{x}{1 + x}

kaosimqu5t

Beginner2022-04-01Added 10 answers

My first instinct would have been to substitute

x = \tan^{2} (θ)

I = \int \frac{x^{n - 2}}{{(1 + x)}^{n}} d x

x = \tan^{2} (θ)

I = \int \frac{\tan^{2 n - 4}}{\sec^{2 n} (θ)} 2 \tan (θ) \sec^{2} (θ) d θ

= 2 \int \frac{\tan^{2 n - 3}}{\sec^{2 n - 2} (θ)} d θ

= \int \sin^{2 n - 3} (θ) \cos (θ) d θ

Setting

\sin (θ) = t

we get

I = 2 \int t^{2 n - 3} d t = 2 \frac{t^{2 n - 2}}{2 n - 2} + C = \frac{{(\sin^{2} (θ))}^{n - 1}}{n - 1} + C

= \frac{1}{n - 1} {(\frac{x}{1 + x})}^{n - 1} + C

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?