What is the integral of (x^2)/((x^3+1)^7) ?

The integral of (x^2)/((x^3+1)^7) is -1/(18(x^3+1)^6)+C

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integral of (x^2)/((x^3+1)^7)

\int \frac{x ^{2}}{( x ^{3} + 1 ) ^{7}} d x

- \frac{1}{18 ( x ^{3} + 1 ) ^{6}} + C

Solution steps

\int \frac{x ^{2}}{( x ^{3} + 1 ) ^{7}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{2}}{( x ^{3} + 1 ) ^{7}} = (x^{2}) (x^{3} + 1)^{- 7}

= \int x^{2} (x^{3} + 1)^{- 7} d x

Apply u-substitution

: \int \frac{1}{3 u ^{7}} d u

= \int \frac{1}{3 u ^{7}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

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

Substitute back

u = x^{3} + 1

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

Simplify

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

= - \frac{1}{18 ( x ^{3} + 1 ) ^{6}}

Add a constant to the solution

= - \frac{1}{18 ( x ^{3} + 1 ) ^{6}} + C

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What is the integral of (x^2)/((x^3+1)^7) ?
The integral of (x^2)/((x^3+1)^7) is -1/(18(x^3+1)^6)+C