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

The integral of (x^4)/(x^2+4x+4) is (x^3)/3-2x^2+12x-32ln|x+2|-(16)/(x+2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{x ^{3}}{3} - 2 x^{2} + 12 x - 32 ln ∣ x + 2 ∣ - \frac{16}{x + 2} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4}}{x ^{2} + 4 x + 4} : x^{2} - 4 x + 12 - \frac{32}{x + 2} + \frac{16}{( x + 2 ) ^{2}}

= \int x^{2} - 4 x + 12 - \frac{32}{x + 2} + \frac{16}{( x + 2 ) ^{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 x^{2} d x - \int 4 x d x + \int 12 d x - \int \frac{32}{x + 2} d x + \int \frac{16}{( x + 2 ) ^{2}} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 12 d x = 12 x

\int \frac{32}{x + 2} d x = 32 ln ∣ x + 2 ∣

\int \frac{16}{( x + 2 ) ^{2}} d x = - \frac{16}{x + 2}

= \frac{x ^{3}}{3} - 2 x^{2} + 12 x - 32 ln ∣ x + 2 ∣ - \frac{16}{x + 2}

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x^{2} + 12 x - 32 ln ∣ x + 2 ∣ - \frac{16}{x + 2} + C

\int 5 sec^{6} (x) d x

\int \frac{x}{3 x + 1} d x

\int_{3}^{8} \frac{x}{x ^{2} - 1} d x

\int cos (x - 2 y) d x

\int_{0}^{\infty} x e^{- \frac{x}{3}} d x

What is the integral of (x^4)/(x^2+4x+4) ?
The integral of (x^4)/(x^2+4x+4) is (x^3)/3-2x^2+12x-32ln|x+2|-(16)/(x+2)+C