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

The integral of (x^3-4x)/((x^2+1)^2) is 1/2 ln|x^2+1|+5/(2(x^2+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3-4x)/((x^2+1)^2)

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

\frac{1}{2} ln x^{2} + 1 + \frac{5}{2 ( x ^{2} + 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} - 4 x}{( x ^{2} + 1 ) ^{2}} : \frac{x}{x ^{2} + 1} - \frac{5 x}{( x ^{2} + 1 ) ^{2}}

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

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

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

= \frac{1}{2} ln x^{2} + 1 - (- \frac{5}{2 ( x ^{2} + 1 )})

Simplify

= \frac{1}{2} ln x^{2} + 1 + \frac{5}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 + \frac{5}{2 ( x ^{2} + 1 )} + C

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

\int \frac{( x + 7 ) ^{6}}{2 x} d x

\int cos (\frac{π}{2} - x) d x

\int \frac{2 - x}{9 x ^{4} + x ^{2}} d x

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

What is the integral of (x^3-4x)/((x^2+1)^2) ?
The integral of (x^3-4x)/((x^2+1)^2) is 1/2 ln|x^2+1|+5/(2(x^2+1))+C