What is the integral of (6x^2)/((x^3-7)^4) ?

The integral of (6x^2)/((x^3-7)^4) is -2/(3(x^3-7)^3)+C

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integral of (6x^2)/((x^3-7)^4)

\int \frac{6 x ^{2}}{( x ^{3} - 7 ) ^{4}} d x

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

Solution steps

\int \frac{6 x ^{2}}{( x ^{3} - 7 ) ^{4}} 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} - 7 ) ^{4}} d x

Apply u-substitution

: \int \frac{u}{2 x ( u ^{\frac{3}{2}} - 7 ) ^{4}} d u

= 6 \cdot \int \frac{1}{3 ( u - 7 ) ^{4}} 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 - 7 ) ^{4}} d u

Apply u-substitution

: \int \frac{1}{v ^{4}} d v

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

: - \frac{1}{3 v ^{3}}

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

Substitute back

= 6 \cdot \frac{1}{3} (- \frac{1}{3 ( x ^{3} - 7 ) ^{3}})

Simplify

6 \cdot \frac{1}{3} (- \frac{1}{3 ( x ^{3} - 7 ) ^{3}}) : - \frac{2}{3 ( x ^{3} - 7 ) ^{3}}

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

Add a constant to the solution

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

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What is the integral of (6x^2)/((x^3-7)^4) ?
The integral of (6x^2)/((x^3-7)^4) is -2/(3(x^3-7)^3)+C