What is the integral of (4x^2+9x+4)/((x^2+1)^2) ?

The integral of (4x^2+9x+4)/((x^2+1)^2) is 4arctan(x)-9/(2(x^2+1))+C

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integral of (4x^2+9x+4)/((x^2+1)^2)

\int \frac{4 x ^{2} + 9 x + 4}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{4 x ^{2} + 9 x + 4}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{4 x ^{2} + 9 x + 4}{( x ^{2} + 1 ) ^{2}} : \frac{4}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{4}{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{4}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

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

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

= 4 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 4 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

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

\int + \frac{1}{x ^{2} + a ^{2}} d x

\frac{d y}{d x} = 4 y e^{x + 4}

(x + y) d y = (x - y) d x

f (x) = x^{4} - 13 x^{2} + 36, at x = - 2

What is the integral of (4x^2+9x+4)/((x^2+1)^2) ?
The integral of (4x^2+9x+4)/((x^2+1)^2) is 4arctan(x)-9/(2(x^2+1))+C