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

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

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

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

- \frac{1}{5} ln 2 - x^{5} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{5} ln ∣ u ∣

Substitute back

u = 2 - x^{5}

= - \frac{1}{5} ln 2 - x^{5}

Add a constant to the solution

= - \frac{1}{5} ln 2 - x^{5} + C

\frac{\partial}{\partial t} (t + sin (t - 1))

\int 3 tan^{3} (2 x) sec^{5} (2 x) d x

x^{3} \cdot y^{^{'}} = x

[(z - 1)^{3} (z + 1)^{2}]

\frac{\partial}{\partial y} (y + (cos (x))^{2})

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