What is the integral of (4x)/((x^2+6)^2) ?

The integral of (4x)/((x^2+6)^2) is -2/(x^2+6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (4x)/((x^2+6)^2)

\int \frac{4 x}{( x ^{2} + 6 ) ^{2}} d x

- \frac{2}{x ^{2} + 6} + C

Solution steps

\int \frac{4 x}{( x ^{2} + 6 ) ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{( x ^{2} + 6 ) ^{2}} d x

Apply u-substitution

= 4 \cdot \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

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

Substitute back

u = x^{2} + 6

= 4 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 6})

Simplify

4 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 6}) : - \frac{2}{x ^{2} + 6}

= - \frac{2}{x ^{2} + 6}

Add a constant to the solution

= - \frac{2}{x ^{2} + 6} + C

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