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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 ln ∣ x ∣ - \frac{1}{x} - ln x^{2} + 1 - arctan (x) + C

Solution steps

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

Take the partial fraction of

\frac{1 + 2 x}{x ^{4} + x ^{2}} : \frac{2}{x} + \frac{1}{x ^{2}} + \frac{- 2 x - 1}{x ^{2} + 1}

= \int \frac{2}{x} + \frac{1}{x ^{2}} + \frac{- 2 x - 1}{x ^{2} + 1} d x

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{2}{x} d x + \int \frac{1}{x ^{2}} d x + \int \frac{- 2 x - 1}{x ^{2} + 1} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

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

Add a constant to the solution

= 2 ln ∣ x ∣ - \frac{1}{x} - ln x^{2} + 1 - arctan (x) + C

\int \frac{( sec ^{2} ( x ) )}{1 + tan ( x )} d x

\int cos (x) sin (x) sin (x) d x

\int \frac{1}{( x + 1 ) 1 - x} d x

\int \frac{1}{1 - 2 t ^{2}} d t

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