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

The integral of x/((3x^2-5)^4) is -1/(18(3x^2-5)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((3x^2-5)^4)

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

- \frac{1}{18 ( 3 x ^{2} - 5 ) ^{3}} + C

Solution steps

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

Apply u-substitution

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

= \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}{3 u ^{3}}

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

Substitute back

u = 3 x^{2} - 5

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

Simplify

\frac{1}{6} (- \frac{1}{3 ( 3 x ^{2} - 5 ) ^{3}}) : - \frac{1}{18 ( 3 x ^{2} - 5 ) ^{3}}

= - \frac{1}{18 ( 3 x ^{2} - 5 ) ^{3}}

Add a constant to the solution

= - \frac{1}{18 ( 3 x ^{2} - 5 ) ^{3}} + C

\int e^{x} - 2 d x

\int (sec (x) - 1)^{2} d x

\int \frac{3 x ^{2} + x - 2}{( x - 1 ) ( x ^{2} + 1 )} d x

\int \frac{1}{sec ^{3} ( 2 x )} d x

\int \frac{1}{x} cos (ln (x)) d x

What is the integral of x/((3x^2-5)^4) ?
The integral of x/((3x^2-5)^4) is -1/(18(3x^2-5)^3)+C