What is the integral of (3y^2)/((y^3-12)^3) ?

The integral of (3y^2)/((y^3-12)^3) is -1/(2(y^3-12)^2)+C

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integral of (3y^2)/((y^3-12)^3)

\int \frac{3 y ^{2}}{( y ^{3} - 12 ) ^{3}} d y

- \frac{1}{2 ( y ^{3} - 12 ) ^{2}} + C

Solution steps

\int \frac{3 y ^{2}}{( y ^{3} - 12 ) ^{3}} d y

Take the constant out:

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

= 3 \cdot \int \frac{y ^{2}}{( y ^{3} - 12 ) ^{3}} d y

Apply u-substitution

: \int \frac{u}{2 y ( u ^{\frac{3}{2}} - 12 ) ^{3}} d u

= 3 \cdot \int \frac{1}{3 ( u - 12 ) ^{3}} 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 - 12 ) ^{3}} d u

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

= 3 \cdot \frac{1}{3} (- \frac{1}{2 ( y ^{3} - 12 ) ^{2}})

Simplify

3 \cdot \frac{1}{3} (- \frac{1}{2 ( y ^{3} - 12 ) ^{2}}) : - \frac{1}{2 ( y ^{3} - 12 ) ^{2}}

= - \frac{1}{2 ( y ^{3} - 12 ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( y ^{3} - 12 ) ^{2}} + C

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What is the integral of (3y^2)/((y^3-12)^3) ?
The integral of (3y^2)/((y^3-12)^3) is -1/(2(y^3-12)^2)+C