What is the integral of (x-4)/((x+2)^2+16) ?

The integral of (x-4)/((x+2)^2+16) is 1/2 ln|x^2+4x+20|-3/2 arctan(1/4 (x+2))+C

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integral of (x-4)/((x+2)^2+16)

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

\frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2)) + C

Solution steps

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{u - 6}{u ^{2} + 16} d u

= \int \frac{u - 6}{u ^{2} + 16} d u

Expand

\frac{u - 6}{u ^{2} + 16} : \frac{u}{u ^{2} + 16} - \frac{6}{u ^{2} + 16}

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{u}{u ^{2} + 16} d u - \int \frac{6}{u ^{2} + 16} d u

\int \frac{u}{u ^{2} + 16} d u = \frac{1}{2} ln u^{2} + 16

\int \frac{6}{u ^{2} + 16} d u = \frac{3}{2} arctan (\frac{u}{4})

= \frac{1}{2} ln u^{2} + 16 - \frac{3}{2} arctan (\frac{u}{4})

Substitute back

u = x + 2

= \frac{1}{2} ln (x + 2)^{2} + 16 - \frac{3}{2} arctan (\frac{x + 2}{4})

Simplify

\frac{1}{2} ln (x + 2)^{2} + 16 - \frac{3}{2} arctan (\frac{x + 2}{4}) : \frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2))

= \frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2)) + C

x \to 0 lim (\frac{e ^{x^{3}} - 1}{x})

\frac{\partial}{\partial x} (2 x - 2 x y)

\int \frac{2 sin ( x )}{1} d x

f (x) = (ln (x))^{4}, at x = 7

\frac{\partial}{\partial x} (3 y^{2} - x)

What is the integral of (x-4)/((x+2)^2+16) ?
The integral of (x-4)/((x+2)^2+16) is 1/2 ln|x^2+4x+20|-3/2 arctan(1/4 (x+2))+C