What is the integral of (6x^2)/((x^3+1)^3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^2)/((x^3+1)^3)

\int \frac{6 x ^{2}}{( x ^{3} + 1 ) ^{3}} d x

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

Solution steps

\int \frac{6 x ^{2}}{( x ^{3} + 1 ) ^{3}} 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 ^{2}}{( x ^{3} + 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{3}} d u

= 6 \cdot \int \frac{1}{3 u ^{3}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 6 \cdot \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{3} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x + 1}{x ^{2} + 2 x - 15} d x

\int sin^{1} (x) d x

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

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

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

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