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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{6} y - 4 x^{3} y^{2})

y^{^{''}} + y = 3 csc^{2} (t)

x \to 0 lim (- \frac{1}{x - 1})

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

\frac{\partial}{\partial x} (\frac{6 y}{x ^{2} + y ^{2} + 5})

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