What is the integral of (3x^2)/((x^3+1)^5) ?

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

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integral of (3x^2)/((x^3+1)^5)

\int \frac{3 x ^{2}}{( x ^{3} + 1 ) ^{5}} d x

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

Solution steps

\int \frac{3 x ^{2}}{( x ^{3} + 1 ) ^{5}} 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 ^{2}}{( x ^{3} + 1 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{5}} d u

= 3 \cdot \int \frac{1}{3 u ^{5}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= 3 \cdot \frac{1}{3} (- \frac{1}{4 u ^{4}})

Substitute back

u = x^{3} + 1

= 3 \cdot \frac{1}{3} (- \frac{1}{4 ( x ^{3} + 1 ) ^{4}})

Simplify

3 \cdot \frac{1}{3} (- \frac{1}{4 ( x ^{3} + 1 ) ^{4}}) : - \frac{1}{4 ( x ^{3} + 1 ) ^{4}}

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

Add a constant to the solution

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

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