What is the integral of 1/(x(x^2+1)^3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x(x^2+1)^3)

\int \frac{1}{x ( x ^{2} + 1 ) ^{3}} d x

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

Solution steps

\int \frac{1}{x ( x ^{2} + 1 ) ^{3}} d x

Take the partial fraction of

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

= \int \frac{1}{x} - \frac{x}{x ^{2} + 1} - \frac{x}{( x ^{2} + 1 ) ^{2}} - \frac{x}{( x ^{2} + 1 ) ^{3}} 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}{x ^{2} + 1} d x - \int \frac{x}{( x ^{2} + 1 ) ^{2}} d x - \int \frac{x}{( x ^{2} + 1 ) ^{3}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

\int \frac{x}{( x ^{2} + 1 ) ^{3}} d x = - \frac{1}{4 ( x ^{2} + 1 ) ^{2}}

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

Simplify

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

Add a constant to the solution

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

\int (8 - x)^{2} d x

\int \frac{1}{x ( 50000 - x )} d x

\int cosh (10 x) d x

\int (\frac{x ^{2} + 1}{x ^{3} + 2 x ^{2} + x}) d x

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

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