What is the integral of (3x^2+x+3)/((x^2+1)^2) ?

The integral of (3x^2+x+3)/((x^2+1)^2) is 3arctan(x)-1/(2(x^2+1))+C

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integral of (3x^2+x+3)/((x^2+1)^2)

\int \frac{3 x ^{2} + x + 3}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{3 x ^{2} + x + 3}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} + x + 3}{( x ^{2} + 1 ) ^{2}} : \frac{3}{x ^{2} + 1} + \frac{x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{3}{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{3}{x ^{2} + 1} d x + \int \frac{x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{3}{x ^{2} + 1} d x = 3 arctan (x)

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

= 3 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 3 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )} + C

\int \frac{1}{( x ^{2} - 8 x ) ^{\frac{3}{2}}} d x

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

\int \frac{x x}{4} d x

\frac{9 x}{8 - tan ( x )}

\frac{\partial}{\partial v} (sin (u))

What is the integral of (3x^2+x+3)/((x^2+1)^2) ?
The integral of (3x^2+x+3)/((x^2+1)^2) is 3arctan(x)-1/(2(x^2+1))+C