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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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

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x \to 0 lim (x^{6} cos (\frac{7}{x}))

\frac{d}{d x} ((16 c x - 4 x^{2})^{\frac{1}{2}})

x \to 0 lim (x^{4} - \frac{x ^{3}}{3})

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