What is the integral of (x^4)/((6+x^5)^2) ?

The integral of (x^4)/((6+x^5)^2) is -1/(5(6+x^5))+C

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integral of (x^4)/((6+x^5)^2)

\int \frac{x ^{4}}{( 6 + x ^{5} ) ^{2}} d x

- \frac{1}{5 ( 6 + x ^{5} )} + C

Solution steps

\int \frac{x ^{4}}{( 6 + 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 = 6 + x^{5}

= \frac{1}{5} (- \frac{1}{6 + x ^{5}})

Simplify

\frac{1}{5} (- \frac{1}{6 + x ^{5}}) : - \frac{1}{5 ( 6 + x ^{5} )}

= - \frac{1}{5 ( 6 + x ^{5} )}

Add a constant to the solution

= - \frac{1}{5 ( 6 + x ^{5} )} + C

\int 4 tan^{3} (sin (x)) cos (x) d x

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 5 y = 2 e^{- 2 x}

\frac{d y}{d t} = 4 y + y^{4}, y (0) = 1

\frac{d}{d x} (\frac{- 24}{x ^{4}})

What is the integral of (x^4)/((6+x^5)^2) ?
The integral of (x^4)/((6+x^5)^2) is -1/(5(6+x^5))+C