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)

Add a constant to the solution

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

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

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

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

\int (x)^{- 1} d x

\int \frac{x ^{4} + 2}{x ^{6} ( x ^{2} + 1 )} d 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