What is the integral of x/((x^2+4)^5) ?

The integral of x/((x^2+4)^5) is -1/(8(x^2+4)^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+4)^5)

\int \frac{x}{( x ^{2} + 4 ) ^{5}} d x

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

Solution steps

\int \frac{x}{( x ^{2} + 4 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{5}} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

\int sin (x) - sin^{2} (x) d x

\int \frac{x - 120}{x ^{2} ( x + 5 )} d x

\int x^{2} e^{- 8 x} d x

\int x arcsin (x) d arcsin (x) d x

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

What is the integral of x/((x^2+4)^5) ?
The integral of x/((x^2+4)^5) is -1/(8(x^2+4)^4)+C