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

The integral of (3x)/((x^2+2)^2) is -3/(2(x^2+2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x)/((x^2+2)^2)

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

- \frac{3}{2 ( x ^{2} + 2 )} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 3 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 2

= 3 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 2})

Simplify

3 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 2}) : - \frac{3}{2 ( x ^{2} + 2 )}

= - \frac{3}{2 ( x ^{2} + 2 )}

Add a constant to the solution

= - \frac{3}{2 ( x ^{2} + 2 )} + C

\int sin (x) cos (6 x) d x

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{7})

arctan (x + y) = y^{2} + \frac{π}{4}, (1, 0)

\int (x^{3} + 5 x^{2} - 2) e^{2 x} d x

y^{^{'}} = x^{2} - y - 2

What is the integral of (3x)/((x^2+2)^2) ?
The integral of (3x)/((x^2+2)^2) is -3/(2(x^2+2))+C