What is the integral of (x+1)/((x+2)^4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+1)/((x+2)^4)

\int \frac{x + 1}{( x + 2 ) ^{4}} d x

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

Solution steps

\int \frac{x + 1}{( x + 2 ) ^{4}} d x

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{4}{5} x d x

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

\int \frac{x ^{2} + 4}{2} d x

\int 3 (\frac{sin ( 2 x )}{sin ( x )}) d x

\int \frac{1}{x ( ln ( x ^{6} ) ) ^{9}} d x

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