What is the integral of ((U-2))/((U^2-3U-1)) ?

The integral of ((U-2))/((U^2-3U-1)) is

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integral of ((U-2))/((U^2-3U-1))

\int \frac{( U - 2 )}{( U ^{2} - 3 U - 1 )} d U

\frac{1}{2} ln 4 U^{2} - 12 U - 4 + \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) - 1 + C

Solution steps

\int \frac{U - 2}{U ^{2} - 3 U - 1} d U

Expand

\frac{U - 2}{U ^{2} - 3 U - 1} : \frac{U}{U ^{2} - 3 U - 1} - \frac{2}{U ^{2} - 3 U - 1}

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} - 3 U - 1} d U - \int \frac{2}{U ^{2} - 3 U - 1} d U

\int \frac{U}{U ^{2} - 3 U - 1} d U = 2 (\frac{1}{4} ln 4 U^{2} - 12 U - 4 - \frac{3}{4 13} (ln \frac{2 U - 3}{13} + 1 - ln \frac{2 U - 3}{13} - 1))

\int \frac{2}{U ^{2} - 3 U - 1} d U = - \frac{2}{13} (ln \frac{2 U - 3}{13} + 1 - ln \frac{2 U - 3}{13} - 1)

= 2 (\frac{1}{4} ln 4 U^{2} - 12 U - 4 - \frac{3}{4 13} (ln \frac{2 U - 3}{13} + 1 - ln \frac{2 U - 3}{13} - 1)) - (- \frac{2}{13} (ln \frac{2 U - 3}{13} + 1 - ln \frac{2 U - 3}{13} - 1))

Simplify

2 (\frac{1}{4} ln 4 U^{2} - 12 U - 4 - \frac{3}{4 13} (ln \frac{2 U - 3}{13} + 1 - ln \frac{2 U - 3}{13} - 1)) - (- \frac{2}{13} (ln \frac{2 U - 3}{13} + 1 - ln \frac{2 U - 3}{13} - 1)) : \frac{1}{2} ln 4 U^{2} - 12 U - 4 + \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) - 1

= \frac{1}{2} ln 4 U^{2} - 12 U - 4 + \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) - 1

Add a constant to the solution

= \frac{1}{2} ln 4 U^{2} - 12 U - 4 + \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) + 1 - \frac{1}{2 13} ln \frac{1}{13} (2 U - 3) - 1 + C

f (x) = 16 - x^{2}, 0, 4

x

\frac{d}{d x} ((x - 5) ln (\frac{x - 5}{10}))

\int \frac{1}{x 64 x ^{2} - 5} d x

\int_{- 1}^{1} (0 - y^{2} + 1) d y

What is the integral of ((U-2))/((U^2-3U-1)) ?
The integral of ((U-2))/((U^2-3U-1)) is