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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^4)/((x^5+1)^2)

\int \frac{x ^{4}}{( x ^{5} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{x ^{4}}{( x ^{5} + 1 ) ^{2}} d x

Apply u-substitution

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

Substitute back

u = x^{5} + 1

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

Simplify

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

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

Add a constant to the solution

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

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