What is the integral of (y-1)/((4+y^2)^2) ?

The integral of (y-1)/((4+y^2)^2) is -1/(2(4+y^2))-1/16 (arctan(y/2)+1/2 sin(2arctan(y/2)))+C

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integral of (y-1)/((4+y^2)^2)

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

- \frac{1}{2 ( 4 + y ^{2} )} - \frac{1}{16} (arctan (\frac{y}{2}) + \frac{1}{2} sin (2 arctan (\frac{y}{2}))) + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{( 4 + y ^{2} ) ^{2}} d y = \frac{1}{16} (arctan (\frac{y}{2}) + \frac{1}{2} sin (2 arctan (\frac{y}{2})))

= - \frac{1}{2 ( 4 + y ^{2} )} - \frac{1}{16} (arctan (\frac{y}{2}) + \frac{1}{2} sin (2 arctan (\frac{y}{2})))

Add a constant to the solution

= - \frac{1}{2 ( 4 + y ^{2} )} - \frac{1}{16} (arctan (\frac{y}{2}) + \frac{1}{2} sin (2 arctan (\frac{y}{2}))) + C

\int (4 x - 3)^{6} d x

x \to 2 + lim (\frac{x + 3}{x - 2})

f (x) = \frac{6}{x}, at x = 1

x \to 2 lim (x^{2} - \frac{1}{x})

\frac{1}{s ^{2} ( s + 1 ) ^{2}}

What is the integral of (y-1)/((4+y^2)^2) ?
The integral of (y-1)/((4+y^2)^2) is -1/(2(4+y^2))-1/16 (arctan(y/2)+1/2 sin(2arctan(y/2)))+C