What is the integral of ((8x^2+3x+8))/((x^2+1)^2) ?

The integral of ((8x^2+3x+8))/((x^2+1)^2) is 8arctan(x)-3/(2(x^2+1))+C

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integral of ((8x^2+3x+8))/((x^2+1)^2)

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

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

Solution steps

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

Take the partial fraction of

\frac{8 x ^{2} + 3 x + 8}{( x ^{2} + 1 ) ^{2}} : \frac{8}{x ^{2} + 1} + \frac{3 x}{( x ^{2} + 1 ) ^{2}}

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

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

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

= 8 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 8 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )} + C

y^{^{'}} = x e^{x}

\int_{0}^{l n (4)} π (e^{- x})^{2} d x

\frac{\partial}{\partial x} ((1 + x^{2})^{z s i n (y)})

\frac{\partial}{\partial y} (y^{2} ln (1 + x^{2} + y^{2}))

\frac{d}{d x} (ln (\frac{x ln ( x ) - 1}{x ln ( x ) + 1}))

What is the integral of ((8x^2+3x+8))/((x^2+1)^2) ?
The integral of ((8x^2+3x+8))/((x^2+1)^2) is 8arctan(x)-3/(2(x^2+1))+C