What is the integral of (x+4)/(x^2+6x-16) ?

The integral of (x+4)/(x^2+6x-16) is ln|x+8|+6(1/10 ln|x-2|-1/10 ln|x+8|)+C

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integral of (x+4)/(x^2+6x-16)

\int \frac{x + 4}{x ^{2} + 6 x - 16} d x

ln ∣ x + 8 ∣ + 6 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x + 8 ∣) + C

Solution steps

\int \frac{x + 4}{x ^{2} + 6 x - 16} d x

Factor

x^{2} + 6 x - 16 : (x - 2) (x + 8)

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

Apply u-substitution

: \int \frac{u + 6}{u ( u + 10 )} d u

= \int \frac{u + 6}{u ( u + 10 )} d u

Expand

\frac{u + 6}{u ( u + 10 )} : \frac{u}{u ( u + 10 )} + \frac{6}{u ( u + 10 )}

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 ( u + 10 )} d u + \int \frac{6}{u ( u + 10 )} d u

\int \frac{u}{u ( u + 10 )} d u = ln ∣ u + 10 ∣

\int \frac{6}{u ( u + 10 )} d u = 6 (\frac{1}{10} ln ∣ u ∣ - \frac{1}{10} ln ∣ u + 10 ∣)

= ln ∣ u + 10 ∣ + 6 (\frac{1}{10} ln ∣ u ∣ - \frac{1}{10} ln ∣ u + 10 ∣)

Substitute back

u = x - 2

= ln ∣ x - 2 + 10 ∣ + 6 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x - 2 + 10 ∣)

Simplify

ln ∣ x - 2 + 10 ∣ + 6 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x - 2 + 10 ∣) : ln ∣ x + 8 ∣ + 6 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x + 8 ∣)

= ln ∣ x + 8 ∣ + 6 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x + 8 ∣)

Add a constant to the solution

= ln ∣ x + 8 ∣ + 6 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x + 8 ∣) + C

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

y = cos (x), at x = - \frac{π}{2}

x \to 0 lim (e^{x} - 2 x^{2})

\frac{\partial}{\partial x} (4 x^{2} + 3 y^{2})

What is the integral of (x+4)/(x^2+6x-16) ?
The integral of (x+4)/(x^2+6x-16) is ln|x+8|+6(1/10 ln|x-2|-1/10 ln|x+8|)+C