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

The integral of (-x+3)/((x^2+1)^2) is (3x)/(2(x^2+1))+3/2 arctan(x)+1/(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)/((x^2+1)^2)

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

\frac{3 x}{2 ( x ^{2} + 1 )} + \frac{3}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )} + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= \frac{3 x}{2 ( x ^{2} + 1 )} + \frac{3}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= \frac{3 x}{2 ( x ^{2} + 1 )} + \frac{3}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )} + C

\int - 1 x^{2} e^{x^{3}} d x

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

\int \frac{x ^{5}}{a ^{3} - x ^{3}} d x

\int \frac{1}{8 u} d u

\int (sec^{2} (2 x)) d x

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