What is the integral of (9x^2+7x+9)/((x^2+1)^2) ?

The integral of (9x^2+7x+9)/((x^2+1)^2) is 9arctan(x)-7/(2(x^2+1))+C

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integral of (9x^2+7x+9)/((x^2+1)^2)

\int \frac{9 x ^{2} + 7 x + 9}{( x ^{2} + 1 ) ^{2}} d x

9 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{9 x ^{2} + 7 x + 9}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{9 x ^{2} + 7 x + 9}{( x ^{2} + 1 ) ^{2}} : \frac{9}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{9}{x ^{2} + 1} + \frac{7 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{7 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{7}{2 ( x ^{2} + 1 )}

= 9 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 9 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

\int \frac{5 - e ^{x}}{e ^{4 x}} d x

\frac{\partial}{\partial x} (ln (x + 9 y + 5 z))

\frac{d}{d x} (4 a e^{4 x})

\int_{- \infty}^{2} x^{2} e^{x} d x

\int \frac{2}{4 x + 1} d x

What is the integral of (9x^2+7x+9)/((x^2+1)^2) ?
The integral of (9x^2+7x+9)/((x^2+1)^2) is 9arctan(x)-7/(2(x^2+1))+C