What is the integral of (x^4)/((2-x^5)^4) ?

The integral of (x^4)/((2-x^5)^4) is 1/(15(2-x^5)^3)+C

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integral of (x^4)/((2-x^5)^4)

\int \frac{x ^{4}}{( 2 - x ^{5} ) ^{4}} d x

\frac{1}{15 ( 2 - x ^{5} ) ^{3}} + C

Solution steps

\int \frac{x ^{4}}{( 2 - x ^{5} ) ^{4}} d x

Apply u-substitution

: \int - \frac{1}{5 u ^{4}} d u

= \int - \frac{1}{5 u ^{4}} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

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

= - \frac{1}{5} (- \frac{1}{3 u ^{3}})

Substitute back

u = 2 - x^{5}

= - \frac{1}{5} (- \frac{1}{3 ( 2 - x ^{5} ) ^{3}})

Simplify

- \frac{1}{5} (- \frac{1}{3 ( 2 - x ^{5} ) ^{3}}) : \frac{1}{15 ( 2 - x ^{5} ) ^{3}}

= \frac{1}{15 ( 2 - x ^{5} ) ^{3}}

Add a constant to the solution

= \frac{1}{15 ( 2 - x ^{5} ) ^{3}} + C

y = \frac{4 x ^{2} + 6 x + 2}{x}

\int e^{2 x} \cdot e^{- x} d x

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

f (x) = 2 x^{3} - 2 x, (1, 0)

\frac{\partial}{\partial x} (x e^{x + y})

What is the integral of (x^4)/((2-x^5)^4) ?
The integral of (x^4)/((2-x^5)^4) is 1/(15(2-x^5)^3)+C