What is the integral of (x^2)/((x^2+4x+5)^2) ?

The integral of (x^2)/((x^2+4x+5)^2) is -xsin(2arctan(x+2))+5/2 arctan(x+2)-5/4 sin(2arctan(x+2))+C

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integral of (x^2)/((x^2+4x+5)^2)

\int \frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} d x

- x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2)) + C

Solution steps

\int \frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} : \frac{1}{x ^{2} + 4 x + 5} + \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{2}}

= \int \frac{1}{x ^{2} + 4 x + 5} + \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{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{1}{x ^{2} + 4 x + 5} d x + \int \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{2}} d x

\int \frac{1}{x ^{2} + 4 x + 5} d x = arctan (x + 2)

\int \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{2}} d x = - x sin (2 arctan (x + 2)) + \frac{3}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

= arctan (x + 2) - x sin (2 arctan (x + 2)) + \frac{3}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

Simplify

arctan (x + 2) - x sin (2 arctan (x + 2)) + \frac{3}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2)) : - x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

= - x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

Add a constant to the solution

= - x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2)) + C

\frac{d}{d x} (\frac{1}{2 x + 5})

\int \frac{cos ^{2} ( x )}{1 - sin ( x )} d x

x \to 2 lim (9 x)

5 x + 6, at x = 2

y = e^{3 x}, at x = \frac{1}{3} ln (3)

What is the integral of (x^2)/((x^2+4x+5)^2) ?
The integral of (x^2)/((x^2+4x+5)^2) is -xsin(2arctan(x+2))+5/2 arctan(x+2)-5/4 sin(2arctan(x+2))+C