What is the integral of 1/([(x-1)^2+1]^2)x-1) ?

The integral of 1/([(x-1)^2+1]^2)x-1) is

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integral of 1/([(x-1)^2+1]^2)x-1)

\int \frac{1}{[ ( x - 1 ) ^{2} + 1 ] ^{2}} d (x - 1) d x

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

Solution steps

\int \frac{1}{[ ( x - 1 ) ^{2} + 1 ] ^{2}} d (x - 1) d x

Take the constant out:

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

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

Simplify

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

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

Expand

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

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

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

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

Add a constant to the solution

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

\int \frac{x ^{3}}{1 + x ^{6}} d x

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What is the integral of 1/([(x-1)^2+1]^2)x-1) ?
The integral of 1/([(x-1)^2+1]^2)x-1) is