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

The integral of x/((x+1)^2(x^2+1)) is 1/(2(x+1))+1/2 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 x/((x+1)^2(x^2+1))

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

\frac{1}{2 ( x + 1 )} + \frac{1}{2} arctan (x) + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

= \frac{1}{2 ( x + 1 )} + \frac{1}{2} arctan (x)

Add a constant to the solution

= \frac{1}{2 ( x + 1 )} + \frac{1}{2} arctan (x) + C

\int_{0}^{0.5} 400 x d x

f (x) = (1 + 2 x)^{11}, (0, 1)

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

\frac{\partial}{\partial x} (\frac{x z}{y ^{2} - z})

sec (x)

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