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

The integral of (x+2)/((x^2+4x)^2) is -1/(8x)+1/(8(x+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+2)/((x^2+4x)^2)

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

- \frac{1}{8 x} + \frac{1}{8 ( x + 4 )} + C

Solution steps

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

Take the partial fraction of

\frac{x + 2}{( x ^{2} + 4 x ) ^{2}} : \frac{1}{8 x ^{2}} - \frac{1}{8 ( x + 4 ) ^{2}}

= \int \frac{1}{8 x ^{2}} - \frac{1}{8 ( x + 4 ) ^{2}} 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}{8 x ^{2}} d x - \int \frac{1}{8 ( x + 4 ) ^{2}} d x

\int \frac{1}{8 x ^{2}} d x = - \frac{1}{8 x}

\int \frac{1}{8 ( x + 4 ) ^{2}} d x = - \frac{1}{8 ( x + 4 )}

= - \frac{1}{8 x} - (- \frac{1}{8 ( x + 4 )})

Simplify

= - \frac{1}{8 x} + \frac{1}{8 ( x + 4 )}

Add a constant to the solution

= - \frac{1}{8 x} + \frac{1}{8 ( x + 4 )} + C

4 x - 8

9 - 4 x, at x = - 1

\int_{1}^{4} - 4 e^{x} d x

\frac{d}{d x} (\frac{x - 8}{7 x ^{2} e ^{x}})

f (x) = (x - 2) (x - 1) (x + 1) (x + 2)

What is the integral of (x+2)/((x^2+4x)^2) ?
The integral of (x+2)/((x^2+4x)^2) is -1/(8x)+1/(8(x+4))+C