What is the integral of (x^4)/(7-x^5) ?

The integral of (x^4)/(7-x^5) is -1/5 ln|7-x^5|+C

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integral of (x^4)/(7-x^5)

\int \frac{x ^{4}}{7 - x ^{5}} d x

- \frac{1}{5} ln 7 - x^{5} + C

Solution steps

\int \frac{x ^{4}}{7 - 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 = 7 - x^{5}

= - \frac{1}{5} ln 7 - x^{5}

Add a constant to the solution

= - \frac{1}{5} ln 7 - x^{5} + C

x \to 1 lim (x^{2})

\frac{\partial}{\partial x} (e^{(x + y + 1)})

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

(x + 3)^{3 x - 2}

y = \frac{1}{x - 5}, (4, - 1)

What is the integral of (x^4)/(7-x^5) ?
The integral of (x^4)/(7-x^5) is -1/5 ln|7-x^5|+C