What is the integral of (x^3+x)/(x^4+2x^2+5) ?

The integral of (x^3+x)/(x^4+2x^2+5) is 1/4 ln|x^4+2x^2+5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3+x)/(x^4+2x^2+5)

\int \frac{x ^{3} + x}{x ^{4} + 2 x ^{2} + 5} d x

\frac{1}{4} ln x^{4} + 2 x^{2} + 5 + C

Solution steps

\int \frac{x ^{3} + x}{x ^{4} + 2 x ^{2} + 5} d x

Apply u-substitution

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 2 x^{2} + 5

= \frac{1}{4} ln x^{4} + 2 x^{2} + 5

Add a constant to the solution

= \frac{1}{4} ln x^{4} + 2 x^{2} + 5 + C

What is the integral of (x^3+x)/(x^4+2x^2+5) ?
The integral of (x^3+x)/(x^4+2x^2+5) is 1/4 ln|x^4+2x^2+5|+C