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

The integral of ((4x^2+4))/((x^2-2x+2)^2) is

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

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

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

Solution steps

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

Factor

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0

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

sin (3 t) cos (3 t)

f (x) = \frac{1}{2 x}, at x = 1

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