What is the integral of (4x^2+7x+4)/((x^2+1)^2) ?

The integral of (4x^2+7x+4)/((x^2+1)^2) is 4arctan(x)-7/(2(x^2+1))+C

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integral of (4x^2+7x+4)/((x^2+1)^2)

\int \frac{4 x ^{2} + 7 x + 4}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{4 x ^{2} + 7 x + 4}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{4 x ^{2} + 7 x + 4}{( x ^{2} + 1 ) ^{2}} : \frac{4}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{4}{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{4}{x ^{2} + 1} d x + \int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{4}{x ^{2} + 1} d x = 4 arctan (x)

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

= 4 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 4 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

\frac{d}{d x} (\frac{x ^{3} - 5 x ^{2}}{x ^{8} - 4})

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

(2 t + 4) i + sin (4 t) j + 2 t^{3} k

x \to 0 + lim (x^{3 x})

y = \frac{2}{x}

What is the integral of (4x^2+7x+4)/((x^2+1)^2) ?
The integral of (4x^2+7x+4)/((x^2+1)^2) is 4arctan(x)-7/(2(x^2+1))+C