What is the integral of (x-8)/((x+4)^2+64) ?

The integral of (x-8)/((x+4)^2+64) is 1/2 ln|x^2+8x+80|-3/2 arctan(1/8 (x+4))+C

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integral of (x-8)/((x+4)^2+64)

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x

\frac{1}{2} ln x^{2} + 8 x + 80 - \frac{3}{2} arctan (\frac{1}{8} (x + 4)) + C

Solution steps

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{u - 12}{u ^{2} + 64} d u

= \int \frac{u - 12}{u ^{2} + 64} d u

Expand

\frac{u - 12}{u ^{2} + 64} : \frac{u}{u ^{2} + 64} - \frac{12}{u ^{2} + 64}

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} + 64} d u - \int \frac{12}{u ^{2} + 64} d u

\int \frac{u}{u ^{2} + 64} d u = \frac{1}{2} ln u^{2} + 64

\int \frac{12}{u ^{2} + 64} d u = \frac{3}{2} arctan (\frac{u}{8})

= \frac{1}{2} ln u^{2} + 64 - \frac{3}{2} arctan (\frac{u}{8})

Substitute back

u = x + 4

= \frac{1}{2} ln (x + 4)^{2} + 64 - \frac{3}{2} arctan (\frac{x + 4}{8})

Simplify

\frac{1}{2} ln (x + 4)^{2} + 64 - \frac{3}{2} arctan (\frac{x + 4}{8}) : \frac{1}{2} ln x^{2} + 8 x + 80 - \frac{3}{2} arctan (\frac{1}{8} (x + 4))

= \frac{1}{2} ln x^{2} + 8 x + 80 - \frac{3}{2} arctan (\frac{1}{8} (x + 4))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 8 x + 80 - \frac{3}{2} arctan (\frac{1}{8} (x + 4)) + C

\frac{d}{d t} (e^{- 4 t})

x \frac{d y}{d x} + 3 x^{2} + 4 y = 0

x \cdot e^{x}

x (x^{3} + 8)

\frac{2 x - 6}{x - 6}

What is the integral of (x-8)/((x+4)^2+64) ?
The integral of (x-8)/((x+4)^2+64) is 1/2 ln|x^2+8x+80|-3/2 arctan(1/8 (x+4))+C