What is the integral of (3x^2+4)/((x^2-2x+2)^2) ?

The integral of (3x^2+4)/((x^2-2x+2)^2) is

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integral of (3x^2+4)/((x^2-2x+2)^2)

\int \frac{3 x ^{2} + 4}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\frac{3 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{1}{2} sin (2 arctan (x - 1)) + 2 arctan (x - 1) - \frac{3 ( ( 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{3 x ^{2} + 4}{( x ^{2} - 2 x + 2 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} + 4}{( x ^{2} - 2 x + 2 ) ^{2}} : \frac{3}{x ^{2} - 2 x + 2} + \frac{6 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}}

= \int \frac{3}{x ^{2} - 2 x + 2} + \frac{6 x - 2}{( x ^{2} - 2 x + 2 ) ^{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} - 2 x + 2} d x + \int \frac{6 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x

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

\int \frac{6 x - 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x = \frac{3 x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{x ^{2} - 2 x + 2} - \frac{3 ( ( x - 1 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) ( x ^{2} - 2 x + 2 ) + 1 )}{x ^{2} - 2 x + 2} - arctan (x - 1) - \frac{1}{2} sin (2 arctan (x - 1))

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

Simplify

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

\int \frac{1}{2 + e ^{x}} d x

\int t + 6 d t

d er i v a t i v e f (x) = \frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}}

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

\frac{d}{d x} (\frac{tan ( 20 x )}{x ^{\frac{2}{3}} + 3 x})

What is the integral of (3x^2+4)/((x^2-2x+2)^2) ?
The integral of (3x^2+4)/((x^2-2x+2)^2) is