What is the integral of (x^4)/((3-x^5)^2) ?

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

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integral of (x^4)/((3-x^5)^2)

\int \frac{x ^{4}}{( 3 - x ^{5} ) ^{2}} d x

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

Solution steps

\int \frac{x ^{4}}{( 3 - x ^{5} ) ^{2}} d x

Apply u-substitution

: \int - \frac{1}{5 u ^{2}} d u

= \int - \frac{1}{5 u ^{2}} 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 ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 3 - x^{5}

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

Simplify

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

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

Add a constant to the solution

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

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y = (ln (x))^{l n (x)}

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