What is the integral of ((x^2+9x+1))/((x^2+1)^2) ?

The integral of ((x^2+9x+1))/((x^2+1)^2) is arctan(x)-9/(2(x^2+1))+C

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integral of ((x^2+9x+1))/((x^2+1)^2)

\int \frac{( x ^{2} + 9 x + 1 )}{( x ^{2} + 1 ) ^{2}} d x

arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{x ^{2} + 9 x + 1}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 9 x + 1}{( x ^{2} + 1 ) ^{2}} : \frac{1}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{1}{x ^{2} + 1} + \frac{9 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{9 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{9}{2 ( x ^{2} + 1 )}

= arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

x^{2} y^{^{''}} + 4 x y^{^{'}} - 4 y = 0, y (1) = 0, y^{^{'}} (1) = 3

\int \frac{225}{x ^{3} - 15 x ^{2}} d x

\int \frac{x ^{2} + 17}{x} d x

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

2 x^{2} + 2 x

What is the integral of ((x^2+9x+1))/((x^2+1)^2) ?
The integral of ((x^2+9x+1))/((x^2+1)^2) is arctan(x)-9/(2(x^2+1))+C