What is the integral of-(2x^2)/((4-5x^3)^7) ?

The integral of-(2x^2)/((4-5x^3)^7) is -1/(45(5x^3-4)^6)+C

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integral of-(2x^2)/((4-5x^3)^7)

\int - \frac{2 x ^{2}}{( 4 - 5 x ^{3} ) ^{7}} d x

- \frac{1}{45 ( 5 x ^{3} - 4 ) ^{6}} + C

Solution steps

\int - \frac{2 x ^{2}}{( 4 - 5 x ^{3} ) ^{7}} d x

Take the constant out:

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

= - 2 \cdot \int \frac{x ^{2}}{( 4 - 5 x ^{3} ) ^{7}} d x

Apply u-substitution

= - 2 \cdot \int \frac{1}{3 ( 4 - 5 u ) ^{7}} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{3} \cdot \int \frac{1}{( 4 - 5 u ) ^{7}} d u

Take the partial fraction of

\frac{1}{( 4 - 5 u ) ^{7}} : - \frac{1}{( 5 u - 4 ) ^{7}}

= - 2 \cdot \frac{1}{3} \cdot \int - \frac{1}{( 5 u - 4 ) ^{7}} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{3} (- \int \frac{1}{( 5 u - 4 ) ^{7}} d u)

Apply u-substitution

= - 2 \cdot \frac{1}{3} (- \int \frac{1}{5 v ^{7}} d v)

Take the constant out:

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

= - 2 \cdot \frac{1}{3} (- \frac{1}{5} \cdot \int \frac{1}{v ^{7}} d v)

Apply the Power Rule

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

Substitute back

= - 2 \cdot \frac{1}{3} (- \frac{1}{5} (- \frac{1}{6 ( 5 x ^{3} - 4 ) ^{6}}))

Simplify

- 2 \cdot \frac{1}{3} (- \frac{1}{5} (- \frac{1}{6 ( 5 x ^{3} - 4 ) ^{6}})) : - \frac{1}{45 ( 5 x ^{3} - 4 ) ^{6}}

= - \frac{1}{45 ( 5 x ^{3} - 4 ) ^{6}}

Add a constant to the solution

= - \frac{1}{45 ( 5 x ^{3} - 4 ) ^{6}} + C

What is the integral of-(2x^2)/((4-5x^3)^7) ?
The integral of-(2x^2)/((4-5x^3)^7) is -1/(45(5x^3-4)^6)+C