What is the integral of ((x^2+5x+1))/((x^2+1)^2) ?

The integral of ((x^2+5x+1))/((x^2+1)^2) is arctan(x)-5/(2(x^2+1))+C

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integral of ((x^2+5x+1))/((x^2+1)^2)

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

arctan (x) - \frac{5}{2 ( x ^{2} + 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 5 x + 1}{( x ^{2} + 1 ) ^{2}} : \frac{1}{x ^{2} + 1} + \frac{5 x}{( x ^{2} + 1 ) ^{2}}

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

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

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

= arctan (x) - \frac{5}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= arctan (x) - \frac{5}{2 ( x ^{2} + 1 )} + C

x (4 - x)^{3}

\frac{d}{d t} (e^{t^{3}})

y = \frac{4}{x} + x + 0.5 - 5 x

t \to 0 lim (\frac{sin ^{2} ( 3 t )}{t})

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

What is the integral of ((x^2+5x+1))/((x^2+1)^2) ?
The integral of ((x^2+5x+1))/((x^2+1)^2) is arctan(x)-5/(2(x^2+1))+C