What is the integral of (x+5)/(x^2-16) ?

The integral of (x+5)/(x^2-16) is 1/2 ln|x^2-16|-5/8 (ln| x/4+1|-ln| x/4-1|)+C

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integral of (x+5)/(x^2-16)

\int \frac{x + 5}{x ^{2} - 16} d x

\frac{1}{2} ln x^{2} - 16 - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

Solution steps

\int \frac{x + 5}{x ^{2} - 16} d x

Expand

\frac{x + 5}{x ^{2} - 16} : \frac{x}{x ^{2} - 16} + \frac{5}{x ^{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{x}{x ^{2} - 16} d x + \int \frac{5}{x ^{2} - 16} d x

\int \frac{x}{x ^{2} - 16} d x = \frac{1}{2} ln x^{2} - 16

\int \frac{5}{x ^{2} - 16} d x = - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

= \frac{1}{2} ln x^{2} - 16 - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 16 - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

x \to 0 lim (x^{x^{6}})

f (x) = \frac{x ^{6}}{8 ^{x}}

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

\frac{d}{d x} (\frac{20 x}{( x ^{2} - 4 ) ^{2}})

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

What is the integral of (x+5)/(x^2-16) ?
The integral of (x+5)/(x^2-16) is 1/2 ln|x^2-16|-5/8 (ln| x/4+1|-ln| x/4-1|)+C