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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

\int sec^{2} (4 t) (3 - tan (4 t))^{3} d t

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

\int \frac{x ^{2} - 2}{x ^{3} - 6 x} d x

\int \frac{5 x + 2}{x ^{2} + 2 x + 10} d x

\int \frac{5 r}{( 1 - r ) ^{4}} d r

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