What is the integral of x/((x+2)^5) ?

The integral of x/((x+2)^5) is -1/(3(x+2)^3)+1/(2(x+2)^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((x+2)^5)

\int \frac{x}{( x + 2 ) ^{5}} d x

- \frac{1}{3 ( x + 2 ) ^{3}} + \frac{1}{2 ( x + 2 ) ^{4}} + C

Solution steps

\int \frac{x}{( x + 2 ) ^{5}} d x

Apply u-substitution

: \int \frac{u - 2}{u ^{5}} d u

= \int \frac{u - 2}{u ^{5}} d u

Expand

\frac{u - 2}{u ^{5}} : \frac{1}{u ^{4}} - \frac{2}{u ^{5}}

= \int \frac{1}{u ^{4}} - \frac{2}{u ^{5}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{4}} d u - \int \frac{2}{u ^{5}} d u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

\int \frac{2}{u ^{5}} d u = - \frac{1}{2 u ^{4}}

= - \frac{1}{3 u ^{3}} - (- \frac{1}{2 u ^{4}})

Substitute back

u = x + 2

= - \frac{1}{3 ( x + 2 ) ^{3}} - (- \frac{1}{2 ( x + 2 ) ^{4}})

Simplify

= - \frac{1}{3 ( x + 2 ) ^{3}} + \frac{1}{2 ( x + 2 ) ^{4}}

Add a constant to the solution

= - \frac{1}{3 ( x + 2 ) ^{3}} + \frac{1}{2 ( x + 2 ) ^{4}} + C

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

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

\int - 3 x^{2} + 6 x d x

\int csc^{2} (2 θ) cot (2 θ) d θ

\int \frac{1}{1 + 2 x + x ^{2}} d x

What is the integral of x/((x+2)^5) ?
The integral of x/((x+2)^5) is -1/(3(x+2)^3)+1/(2(x+2)^4)+C