What is the integral of (x+3)/(x^2+6x^4) ?

The integral of (x+3)/(x^2+6x^4) is ln|x|-3/x-1/2 ln|6x^2+1|-3sqrt(6)arctan(sqrt(6)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)/(x^2+6x^4)

\int \frac{x + 3}{x ^{2} + 6 x ^{4}} d x

ln ∣ x ∣ - \frac{3}{x} - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x) + C

Solution steps

\int \frac{x + 3}{x ^{2} + 6 x ^{4}} d x

Take the partial fraction of

\frac{x + 3}{x ^{2} + 6 x ^{4}} : \frac{1}{x} + \frac{3}{x ^{2}} + \frac{- 6 x - 18}{6 x ^{2} + 1}

= \int \frac{1}{x} + \frac{3}{x ^{2}} + \frac{- 6 x - 18}{6 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{1}{x} d x + \int \frac{3}{x ^{2}} d x + \int \frac{- 6 x - 18}{6 x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{x ^{2}} d x = - \frac{3}{x}

\int \frac{- 6 x - 18}{6 x ^{2} + 1} d x = - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x)

= ln ∣ x ∣ - \frac{3}{x} - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x)

Add a constant to the solution

= ln ∣ x ∣ - \frac{3}{x} - \frac{1}{2} ln 6 x^{2} + 1 - 36 arctan (6 x) + C

\int 10 x ln (3 x) d x

\int cos (θ) sin (θ) d θ

\int_{- \infty}^{\infty} 57 cos (π t) d t

y^{^{'}} = 2 x y^{2} + 3 x^{2} y^{2}, y (1) = - 1

\frac{( x ^{2} - x + 1 )}{x + 1}

What is the integral of (x+3)/(x^2+6x^4) ?
The integral of (x+3)/(x^2+6x^4) is ln|x|-3/x-1/2 ln|6x^2+1|-3sqrt(6)arctan(sqrt(6)x)+C