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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int \frac{25 - 16 x ^{2}}{x} d x

\int \frac{1}{1 - v ^{2}} d v

\int \frac{1}{x ^{3} ( ( x ^{2} - 9 ) )} d x

d er i v a t i v e \frac{1}{u ^{\frac{1}{2}}}

t an g e n t f (x) = 2 x^{2} - 3 x, at x = 4

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