What is the integral of (2x^2+3)/((x^2+1)^2) ?

The integral of (2x^2+3)/((x^2+1)^2) is 5/2 arctan(x)+x/(2(1+x^2))+C

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integral of (2x^2+3)/((x^2+1)^2)

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

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

Solution steps

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

Take the partial fraction of

\frac{2 x ^{2} + 3}{( x ^{2} + 1 ) ^{2}} : \frac{2}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{2}}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

(x + y) \frac{d y}{d x} = x - y

\int sec^{4} (x) tan^{3} (x) d x

\frac{1 - 2 s}{5 s ^{2} + 8 s + 10}

y^{^{'}} = t + t y^{2}

f (x) = x + x, at x = 4

What is the integral of (2x^2+3)/((x^2+1)^2) ?
The integral of (2x^2+3)/((x^2+1)^2) is 5/2 arctan(x)+x/(2(1+x^2))+C