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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{3 u ^{2}} 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 ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = x^{3} - 4

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

Simplify

\frac{1}{3} (- \frac{1}{x ^{3} - 4}) : - \frac{1}{3 ( x ^{3} - 4 )}

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

Add a constant to the solution

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

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\int e^{x} (x^{2} + 2) d x

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