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

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

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

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

- \frac{1}{6 ( 2 x ^{3} - 5 ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{2 x ( 2 u ^{\frac{3}{2}} - 5 ) ^{3}} d u

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

Apply u-substitution

: \int \frac{1}{2 v ^{3}} d v

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{2 v ^{3}} 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} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{3}} d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{6 ( 2 x ^{3} - 5 ) ^{2}} + C

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