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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

f (x) = 2 e^{x}, at x = 2

\int 9 (3 x + 2)^{2} d x

\int sin (22 x) cos (15 x) d x

y = 2 x - x^{2}, y = - x

y = sin (h) 3 x

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