What is the integral of (6x)/((5+3x^2)^4) ?

The integral of (6x)/((5+3x^2)^4) is -1/(3(5+3x^2)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x)/((5+3x^2)^4)

\int \frac{6 x}{( 5 + 3 x ^{2} ) ^{4}} d x

- \frac{1}{3 ( 5 + 3 x ^{2} ) ^{3}} + C

Solution steps

\int \frac{6 x}{( 5 + 3 x ^{2} ) ^{4}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{( 5 + 3 x ^{2} ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{6 u ^{4}} d u

= 6 \cdot \int \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 6 \cdot \frac{1}{6} (- \frac{1}{3 u ^{3}})

Substitute back

u = 5 + 3 x^{2}

= 6 \cdot \frac{1}{6} (- \frac{1}{3 ( 5 + 3 x ^{2} ) ^{3}})

Simplify

6 \cdot \frac{1}{6} (- \frac{1}{3 ( 5 + 3 x ^{2} ) ^{3}}) : - \frac{1}{3 ( 5 + 3 x ^{2} ) ^{3}}

= - \frac{1}{3 ( 5 + 3 x ^{2} ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( 5 + 3 x ^{2} ) ^{3}} + C

x, 5 x, - x + 4

\frac{d y}{d x}, y = x^{9}

f (x) = x^{2} - 3 x + 7

\frac{d}{d x} (x^{4} - x)

\frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}}

What is the integral of (6x)/((5+3x^2)^4) ?
The integral of (6x)/((5+3x^2)^4) is -1/(3(5+3x^2)^3)+C