What is the integral of (3x^2+7x+3)/((x^2+1)^2) ?

The integral of (3x^2+7x+3)/((x^2+1)^2) is 3arctan(x)-7/(2(x^2+1))+C

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integral of (3x^2+7x+3)/((x^2+1)^2)

\int \frac{3 x ^{2} + 7 x + 3}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{3 x ^{2} + 7 x + 3}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} + 7 x + 3}{( x ^{2} + 1 ) ^{2}} : \frac{3}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{3}{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{3}{x ^{2} + 1} d x + \int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x

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

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

= 3 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 3 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

\frac{d}{d x} (8^{- x})

x \to 0 lim (\frac{tan ( 3 x )}{x})

\int cos (2 - \frac{x}{3}) d x

x \to - 2 lim (\frac{x - 2}{x ^{3} + 8})

\frac{\partial}{\partial θ} (r (θ) \cdot sin (θ))

What is the integral of (3x^2+7x+3)/((x^2+1)^2) ?
The integral of (3x^2+7x+3)/((x^2+1)^2) is 3arctan(x)-7/(2(x^2+1))+C