What is the integral of (6x^2+x+6)/((x^2+1)^2) ?

The integral of (6x^2+x+6)/((x^2+1)^2) is 6arctan(x)-1/(2(x^2+1))+C

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integral of (6x^2+x+6)/((x^2+1)^2)

\int \frac{6 x ^{2} + x + 6}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{6 x ^{2} + x + 6}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{6 x ^{2} + x + 6}{( x ^{2} + 1 ) ^{2}} : \frac{6}{x ^{2} + 1} + \frac{x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{6}{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{6}{x ^{2} + 1} d x + \int \frac{x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

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

= 6 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 6 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )} + C

ln (x - 4)

\frac{d}{d x} (ln (sin (4 x^{3} + 6 x^{2} + 8)))

\int \frac{2 x + 5}{x ^{2}} d x

\int \frac{1}{( x ^{2} + 4 x + 5 )} d x

\frac{d}{d x} (4 - \frac{600}{x ^{2}})

What is the integral of (6x^2+x+6)/((x^2+1)^2) ?
The integral of (6x^2+x+6)/((x^2+1)^2) is 6arctan(x)-1/(2(x^2+1))+C