What is the integral of (x^5)/((4-x^6)^4) ?

The integral of (x^5)/((4-x^6)^4) is 1/(18(4-x^6)^3)+C

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integral of (x^5)/((4-x^6)^4)

\int \frac{x ^{5}}{( 4 - x ^{6} ) ^{4}} d x

\frac{1}{18 ( 4 - x ^{6} ) ^{3}} + C

Solution steps

\int \frac{x ^{5}}{( 4 - x ^{6} ) ^{4}} d x

Apply u-substitution

: \int - \frac{1}{6 u ^{4}} d u

= \int - \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

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

Substitute back

u = 4 - x^{6}

= - \frac{1}{6} (- \frac{1}{3 ( 4 - x ^{6} ) ^{3}})

Simplify

- \frac{1}{6} (- \frac{1}{3 ( 4 - x ^{6} ) ^{3}}) : \frac{1}{18 ( 4 - x ^{6} ) ^{3}}

= \frac{1}{18 ( 4 - x ^{6} ) ^{3}}

Add a constant to the solution

= \frac{1}{18 ( 4 - x ^{6} ) ^{3}} + C

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

f (x) = \frac{x}{x ^{2} + 16}, at x = 0

\int \frac{x ^{2}}{( 15 + 4 x - 4 x ^{2} ) ^{\frac{3}{2}}} d x

\int (x^{2} e^{x^{3}} + x^{2} sin (y)) d x

\int (\frac{x - 5}{x ^{2} - x - 6}) d x

What is the integral of (x^5)/((4-x^6)^4) ?
The integral of (x^5)/((4-x^6)^4) is 1/(18(4-x^6)^3)+C