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

The integral of (x^2)/((x^3-1)^4) is -1/(9(x^3-1)^3)+C

integral of (x^2)/((x^3-1)^4)

\int \frac{x ^{2}}{( x ^{3} - 1 ) ^{4}} d x

- \frac{1}{9 ( x ^{3} - 1 ) ^{3}} + C

Solution steps

\int \frac{x ^{2}}{( x ^{3} - 1 ) ^{4}} d x

Apply u-substitution

= \int \frac{1}{3 ( u - 1 ) ^{4}} 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 - 1 ) ^{4}} d u

Apply u-substitution

= \frac{1}{3} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

Substitute back

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

Simplify

\frac{1}{3} (- \frac{1}{3 ( x ^{3} - 1 ) ^{3}}) : - \frac{1}{9 ( x ^{3} - 1 ) ^{3}}

= - \frac{1}{9 ( x ^{3} - 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{9 ( x ^{3} - 1 ) ^{3}} + C

\int 9 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

\frac{d}{d x} (\frac{x ^{3}}{x ^{2} + 1})

x \to 0 lim (\frac{e ^{- 5 x} - 1}{x})

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

t an g e n t f (x) = 2 x^{2} + 3 x, at x = - 2

What is the integral of (x^2)/((x^3-1)^4) ?
The integral of (x^2)/((x^3-1)^4) is -1/(9(x^3-1)^3)+C