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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Simplify

\frac{x ^{3} + 2 x}{x ^{4} + 4 x ^{2}} : \frac{x ^{2} + 2}{x ( x ^{2} + 4 )}

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

Expand

\frac{x ^{2} + 2}{x ( x ^{2} + 4 )} : \frac{x ^{2}}{x ( x ^{2} + 4 )} + \frac{2}{x ( x ^{2} + 4 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{2}{x ( x ^{2} + 4 )} d x = 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{1}{2} ln x^{2} + 4 + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

\frac{1}{2} ln x^{2} + 4 + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{1}{4} ln x^{2} + 4 + \frac{1}{2} ln ∣ x ∣

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

Add a constant to the solution

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

\int (2 x - 3) (4 x^{2} + 6 x + 9) d x

\int \frac{x ^{3}}{x ^{4} - 4} d x

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

\int \frac{b}{1 - y} d y

\int \frac{1}{tan ^{4} ( x )} d x

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