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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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