What is the integral of (2x^2+7x+2)/((x^2+1)^2) ?

The integral of (2x^2+7x+2)/((x^2+1)^2) is 2arctan(x)-7/(2(x^2+1))+C

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integral of (2x^2+7x+2)/((x^2+1)^2)

\int \frac{2 x ^{2} + 7 x + 2}{( x ^{2} + 1 ) ^{2}} d x

2 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{2 x ^{2} + 7 x + 2}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{2 x ^{2} + 7 x + 2}{( x ^{2} + 1 ) ^{2}} : \frac{2}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{2}{x ^{2} + 1} + \frac{7 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{7 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{7}{2 ( x ^{2} + 1 )}

= 2 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 2 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

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

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

\int \frac{arcsin ( x )}{x} d x

y^{^{''}} + 16 y^{^{'}} + 80 y = 0

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

What is the integral of (2x^2+7x+2)/((x^2+1)^2) ?
The integral of (2x^2+7x+2)/((x^2+1)^2) is 2arctan(x)-7/(2(x^2+1))+C