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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

(x^{6} + y^{6}) d x + 6 x y^{5} d y = 0

f (t) = cos^{2} (t)

\frac{d}{d x} (ln (1 - 4 x))

x + 10

\frac{s + 3.3}{s + 1}

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