What is the integral of (5x^2+9x+5)/((x^2+1)^2) ?

The integral of (5x^2+9x+5)/((x^2+1)^2) is 5arctan(x)-9/(2(x^2+1))+C

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integral of (5x^2+9x+5)/((x^2+1)^2)

\int \frac{5 x ^{2} + 9 x + 5}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{5 x ^{2} + 9 x + 5}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{5 x ^{2} + 9 x + 5}{( x ^{2} + 1 ) ^{2}} : \frac{5}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{5}{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{5}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

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

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

= 5 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 5 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

\frac{d}{d a} (sin (a) - sin (2 a) + sin (3 a))

x \to 0 - lim (x cot (x))

y = \frac{7}{x ^{3}}

\frac{d}{d x} (\frac{1}{π - x})

\frac{y ^{^{'}}}{y - y ^{2}} = \frac{t}{( 1 + t ^{2} )}

What is the integral of (5x^2+9x+5)/((x^2+1)^2) ?
The integral of (5x^2+9x+5)/((x^2+1)^2) is 5arctan(x)-9/(2(x^2+1))+C