What is the integral of (5y^4)/(y^5-13) ?

The integral of (5y^4)/(y^5-13) is ln|y^5-13|+C

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integral of (5y^4)/(y^5-13)

\int \frac{5 y ^{4}}{y ^{5} - 13} d y

ln y^{5} - 13 + C

Solution steps

\int \frac{5 y ^{4}}{y ^{5} - 13} d y

Take the constant out:

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

= 5 \cdot \int \frac{y ^{4}}{y ^{5} - 13} d y

Apply u-substitution

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

= 5 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{5} ln ∣ u ∣

Substitute back

u = y^{5} - 13

= 5 \cdot \frac{1}{5} ln y^{5} - 13

Simplify

5 \cdot \frac{1}{5} ln y^{5} - 13 : ln y^{5} - 13

= ln y^{5} - 13

Add a constant to the solution

= ln y^{5} - 13 + C

\frac{50 s}{s ^{2} + 25}

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

\int_{- 2}^{5} - x^{3} + 9 x d x

\frac{d}{d x} (4 cot (x))

f (x) = - 5 e^{- 2.5 x^{2}} x

What is the integral of (5y^4)/(y^5-13) ?
The integral of (5y^4)/(y^5-13) is ln|y^5-13|+C