What is the integral of (x^3)/((x+2)^4) ?

The integral of (x^3)/((x+2)^4) is ln|x+2|+6/(x+2)-6/((x+2)^2)+8/(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^3)/((x+2)^4)

integral

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

ln ∣ x + 2 ∣ + \frac{6}{x + 2} - \frac{6}{( x + 2 ) ^{2}} + \frac{8}{3 ( x + 2 ) ^{3}} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3}}{( x + 2 ) ^{4}} : \frac{1}{x + 2} - \frac{6}{( x + 2 ) ^{2}} + \frac{12}{( x + 2 ) ^{3}} - \frac{8}{( x + 2 ) ^{4}}

= \int \frac{1}{x + 2} - \frac{6}{( x + 2 ) ^{2}} + \frac{12}{( x + 2 ) ^{3}} - \frac{8}{( 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} d x - \int \frac{6}{( x + 2 ) ^{2}} d x + \int \frac{12}{( x + 2 ) ^{3}} d x - \int \frac{8}{( x + 2 ) ^{4}} d x

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

\int \frac{6}{( x + 2 ) ^{2}} d x = - \frac{6}{x + 2}

\int \frac{12}{( x + 2 ) ^{3}} d x = - \frac{6}{( x + 2 ) ^{2}}

\int \frac{8}{( x + 2 ) ^{4}} d x = - \frac{8}{3 ( x + 2 ) ^{3}}

= ln ∣ x + 2 ∣ - (- \frac{6}{x + 2}) - \frac{6}{( x + 2 ) ^{2}} - (- \frac{8}{3 ( x + 2 ) ^{3}})

Simplify

= ln ∣ x + 2 ∣ + \frac{6}{x + 2} - \frac{6}{( x + 2 ) ^{2}} + \frac{8}{3 ( x + 2 ) ^{3}}

Add a constant to the solution

= ln ∣ x + 2 ∣ + \frac{6}{x + 2} - \frac{6}{( x + 2 ) ^{2}} + \frac{8}{3 ( x + 2 ) ^{3}} + C

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

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

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

\frac{sin ^{7} ( x )}{cos ^{2} ( x )}

x^{2} - 2 x + 1

What is the integral of (x^3)/((x+2)^4) ?
The integral of (x^3)/((x+2)^4) is ln|x+2|+6/(x+2)-6/((x+2)^2)+8/(3(x+2)^3)+C