What is the integral of (8x^2+9x+8)/((x^2+1)^2) ?

The integral of (8x^2+9x+8)/((x^2+1)^2) is 8arctan(x)-9/(2(x^2+1))+C

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integral of (8x^2+9x+8)/((x^2+1)^2)

\int \frac{8 x ^{2} + 9 x + 8}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{8 x ^{2} + 9 x + 8}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{8 x ^{2} + 9 x + 8}{( x ^{2} + 1 ) ^{2}} : \frac{8}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{8}{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{8}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x)

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

= 8 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 8 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

n = 0 \sum \infty (\frac{3}{7})^{n}

\frac{d y}{d x} = 9 x^{3}

\frac{d}{d x} (cot (π) x \cdot sec (π) x)

5 x + 6

f (x) = ln (x + x^{2} - 12)

What is the integral of (8x^2+9x+8)/((x^2+1)^2) ?
The integral of (8x^2+9x+8)/((x^2+1)^2) is 8arctan(x)-9/(2(x^2+1))+C