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

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

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

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

2 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )} + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= 2 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 2 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )} + C

1 + e^{4 t}

x \to - 4 lim (8 x + 3)

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

x \frac{d y}{d x} + 4 y = 2 - \frac{7}{x}

\int (x^{3} + 4 x - 3) d x

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