What is the integral of x/((3x^2-4)^4) ?

The integral of x/((3x^2-4)^4) is -1/(18(3x^2-4)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x/((3x^2-4)^4)

\int \frac{x}{( 3 x ^{2} - 4 ) ^{4}} d x

- \frac{1}{18 ( 3 x ^{2} - 4 ) ^{3}} + C

Solution steps

\int \frac{x}{( 3 x ^{2} - 4 ) ^{4}} d x

Apply u-substitution

= \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

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

Apply the Power Rule

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

Substitute back

u = 3 x^{2} - 4

= \frac{1}{6} (- \frac{1}{3 ( 3 x ^{2} - 4 ) ^{3}})

Simplify

\frac{1}{6} (- \frac{1}{3 ( 3 x ^{2} - 4 ) ^{3}}) : - \frac{1}{18 ( 3 x ^{2} - 4 ) ^{3}}

= - \frac{1}{18 ( 3 x ^{2} - 4 ) ^{3}}

Add a constant to the solution

= - \frac{1}{18 ( 3 x ^{2} - 4 ) ^{3}} + C

\int (4 x^{2} - 3 cos (x)) d x

\int sin (10 x) cos (5 x) d x

f (x) = e^{l n (s i n (x^{2}))} 2 x + 1

\int e^{4 x} cos (6 x) d x

\frac{d x}{d t} = 3 x t^{2}

What is the integral of x/((3x^2-4)^4) ?
The integral of x/((3x^2-4)^4) is -1/(18(3x^2-4)^3)+C