What is the integral of (4x^2+x+4)/((x^2+1)^2) ?

The integral of (4x^2+x+4)/((x^2+1)^2) is 4arctan(x)-1/(2(x^2+1))+C

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integral of (4x^2+x+4)/((x^2+1)^2)

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

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

Solution steps

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

Take the partial fraction of

\frac{4 x ^{2} + x + 4}{( x ^{2} + 1 ) ^{2}} : \frac{4}{x ^{2} + 1} + \frac{x}{( x ^{2} + 1 ) ^{2}}

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

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

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

= 4 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 4 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )} + C

\int_{a}^{b} (\frac{1}{x ^{3}} + 3 x) d x

\int - \frac{1}{1 + sin ( x )} d x

x y^{^{'}} - 2 y = x^{4} - 4

\frac{( 3 y ^{2} - t ^{2} )}{( y ^{5} )} \cdot y^{^{'}} + \frac{t}{( 2 y ^{4} )} = 0, y (1) = 3

\frac{e ^{x} - 1}{x}

What is the integral of (4x^2+x+4)/((x^2+1)^2) ?
The integral of (4x^2+x+4)/((x^2+1)^2) is 4arctan(x)-1/(2(x^2+1))+C