What is the integral of (x-6)/((x+3)^2+36) ?

The integral of (x-6)/((x+3)^2+36) is 1/2 ln|x^2+6x+45|-3/2 arctan(1/6 (x+3))+C

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integral of (x-6)/((x+3)^2+36)

\int \frac{x - 6}{( x + 3 ) ^{2} + 36} d x

\frac{1}{2} ln x^{2} + 6 x + 45 - \frac{3}{2} arctan (\frac{1}{6} (x + 3)) + C

Solution steps

\int \frac{x - 6}{( x + 3 ) ^{2} + 36} d x

Apply u-substitution

= \int \frac{u - 9}{u ^{2} + 36} d u

Expand

\frac{u - 9}{u ^{2} + 36} : \frac{u}{u ^{2} + 36} - \frac{9}{u ^{2} + 36}

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} + 36} d u - \int \frac{9}{u ^{2} + 36} d u

\int \frac{u}{u ^{2} + 36} d u = \frac{1}{2} ln u^{2} + 36

\int \frac{9}{u ^{2} + 36} d u = \frac{3}{2} arctan (\frac{u}{6})

= \frac{1}{2} ln u^{2} + 36 - \frac{3}{2} arctan (\frac{u}{6})

Substitute back

u = x + 3

= \frac{1}{2} ln (x + 3)^{2} + 36 - \frac{3}{2} arctan (\frac{x + 3}{6})

Simplify

\frac{1}{2} ln (x + 3)^{2} + 36 - \frac{3}{2} arctan (\frac{x + 3}{6}) : \frac{1}{2} ln x^{2} + 6 x + 45 - \frac{3}{2} arctan (\frac{1}{6} (x + 3))

= \frac{1}{2} ln x^{2} + 6 x + 45 - \frac{3}{2} arctan (\frac{1}{6} (x + 3))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 45 - \frac{3}{2} arctan (\frac{1}{6} (x + 3)) + C

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\frac{d}{d x} ((sin (x)) (cos (x)))

\frac{d y}{d x}, x y = 15

\frac{\partial}{\partial x} (2 sin (x^{2} + y^{2}))

What is the integral of (x-6)/((x+3)^2+36) ?
The integral of (x-6)/((x+3)^2+36) is 1/2 ln|x^2+6x+45|-3/2 arctan(1/6 (x+3))+C