What is the integral of ((2x^2+9x+2))/((x^2+1)^2) ?

The integral of ((2x^2+9x+2))/((x^2+1)^2) is 2arctan(x)-9/(2(x^2+1))+C

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integral of ((2x^2+9x+2))/((x^2+1)^2)

\int \frac{( 2 x ^{2} + 9 x + 2 )}{( x ^{2} + 1 ) ^{2}} d x

2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} : \frac{2}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{2}{x ^{2} + 1} + \frac{9 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{2}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{9}{2 ( x ^{2} + 1 )}

= 2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

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

\int \frac{7}{3} d x

\frac{1}{x}, at x = 4

\frac{d}{d x} (cot (9) (1 - 2 x^{2}))

n \to 0 lim (n sin (\frac{1}{n}))

What is the integral of ((2x^2+9x+2))/((x^2+1)^2) ?
The integral of ((2x^2+9x+2))/((x^2+1)^2) is 2arctan(x)-9/(2(x^2+1))+C