What is the integral of (x-1)/((x^2+2x+3)^2) ?

The integral of (x-1)/((x^2+2x+3)^2) is

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integral of (x-1)/((x^2+2x+3)^2)

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

\frac{1}{8 2} (2 arctan (\frac{x + 1}{2}) + sin (2 arctan (\frac{x + 1}{2}))) (x - 1) - \frac{1}{4} \frac{( x + 1 ) ( arctan ( \frac{1}{2} ( x + 1 ) ) ( x ^{2} + 2 x + 3 ) + 2 ( x + 1 ) )}{2 ( x ^{2} + 2 x + 3 )} + \frac{2}{x ^{2} + 2 x + 3} + C

Solution steps

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

Complete the square

x^{2} + 2 x + 3 : (x + 1)^{2} + 2

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

Apply u-substitution

: \int \frac{u - 2}{( u ^{2} + 2 ) ^{2}} d u

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

Apply Integration By Parts

: \frac{1}{8 2} (2 arctan (\frac{u}{2}) + sin (2 arctan (\frac{u}{2}))) (u - 2) - \int \frac{1}{4 2} (arctan (\frac{u}{2}) + \frac{1}{2} sin (2 arctan (\frac{u}{2}))) d u

= \frac{1}{8 2} (2 arctan (\frac{u}{2}) + sin (2 arctan (\frac{u}{2}))) (u - 2) - \int \frac{1}{4 2} (arctan (\frac{u}{2}) + \frac{1}{2} sin (2 arctan (\frac{u}{2}))) d u

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

= \frac{1}{8 2} (2 arctan (\frac{u}{2}) + sin (2 arctan (\frac{u}{2}))) (u - 2) - \frac{1}{4} \frac{u ( arctan ( \frac{u}{2} ) ( 2 + u ^{2} ) + 2 u )}{2 ( 2 + u ^{2} )} + \frac{2}{u ^{2} + 2}

Substitute back

u = x + 1

= \frac{1}{8 2} (2 arctan (\frac{x + 1}{2}) + sin (2 arctan (\frac{x + 1}{2}))) (x + 1 - 2) - \frac{1}{4} \frac{( x + 1 ) ( arctan ( \frac{x + 1}{2} ) ( 2 + ( x + 1 ) ^{2} ) + 2 ( x + 1 ) )}{2 ( 2 + ( x + 1 ) ^{2} )} + \frac{2}{( x + 1 ) ^{2} + 2}

Simplify

\frac{1}{8 2} (2 arctan (\frac{x + 1}{2}) + sin (2 arctan (\frac{x + 1}{2}))) (x + 1 - 2) - \frac{1}{4} \frac{( x + 1 ) ( arctan ( \frac{x + 1}{2} ) ( 2 + ( x + 1 ) ^{2} ) + 2 ( x + 1 ) )}{2 ( 2 + ( x + 1 ) ^{2} )} + \frac{2}{( x + 1 ) ^{2} + 2} : \frac{1}{8 2} (2 arctan (\frac{x + 1}{2}) + sin (2 arctan (\frac{x + 1}{2}))) (x - 1) - \frac{1}{4} \frac{( x + 1 ) ( arctan ( \frac{1}{2} ( x + 1 ) ) ( x ^{2} + 2 x + 3 ) + 2 ( x + 1 ) )}{2 ( x ^{2} + 2 x + 3 )} + \frac{2}{x ^{2} + 2 x + 3}

= \frac{1}{8 2} (2 arctan (\frac{x + 1}{2}) + sin (2 arctan (\frac{x + 1}{2}))) (x - 1) - \frac{1}{4} \frac{( x + 1 ) ( arctan ( \frac{1}{2} ( x + 1 ) ) ( x ^{2} + 2 x + 3 ) + 2 ( x + 1 ) )}{2 ( x ^{2} + 2 x + 3 )} + \frac{2}{x ^{2} + 2 x + 3}

Add a constant to the solution

= \frac{1}{8 2} (2 arctan (\frac{x + 1}{2}) + sin (2 arctan (\frac{x + 1}{2}))) (x - 1) - \frac{1}{4} \frac{( x + 1 ) ( arctan ( \frac{1}{2} ( x + 1 ) ) ( x ^{2} + 2 x + 3 ) + 2 ( x + 1 ) )}{2 ( x ^{2} + 2 x + 3 )} + \frac{2}{x ^{2} + 2 x + 3} + C

f (x) = \frac{1}{x}, at x = 0.501

\int 4 x + 4 sin (x) d x

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

\frac{d}{d x} (6 arcsin (x^{4}))

What is the integral of (x-1)/((x^2+2x+3)^2) ?
The integral of (x-1)/((x^2+2x+3)^2) is