What is the integral of x/((4x^2+9)^2) ?

The integral of x/((4x^2+9)^2) is -1/(8(4x^2+9))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x/((4x^2+9)^2)

\int \frac{x}{( 4 x ^{2} + 9 ) ^{2}} d x

- \frac{1}{8 ( 4 x ^{2} + 9 )} + C

Solution steps

\int \frac{x}{( 4 x ^{2} + 9 ) ^{2}} d x

Apply u-substitution

= \int \frac{1}{8 u ^{2}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= \frac{1}{8} (- \frac{1}{u})

Substitute back

u = 4 x^{2} + 9

= \frac{1}{8} (- \frac{1}{4 x ^{2} + 9})

Simplify

\frac{1}{8} (- \frac{1}{4 x ^{2} + 9}) : - \frac{1}{8 ( 4 x ^{2} + 9 )}

= - \frac{1}{8 ( 4 x ^{2} + 9 )}

Add a constant to the solution

= - \frac{1}{8 ( 4 x ^{2} + 9 )} + C

What is the integral of x/((4x^2+9)^2) ?
The integral of x/((4x^2+9)^2) is -1/(8(4x^2+9))+C