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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{1}{x ^{10}}

f (x) = x + 7, (9, 4)

\frac{d x}{d t} = \frac{2 x ^{2} + t 7 t ^{2} + x ^{2}}{2 t x}

f (x) = cos (\frac{e ^{x}}{x ^{4} + 1})

7 t \cdot \frac{d y}{d t} + y = t

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