What is the integral from-3 to 3 of (20)/(x^2-6x-91) ?

The integral from-3 to 3 of (20)/(x^2-6x-91) is -ln(5/2)+ln(5/8)

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integral from-3 to 3 of (20)/(x^2-6x-91)

\int_{- 3}^{3} \frac{20}{x ^{2} - 6 x - 91} d x

- ln (\frac{5}{2}) + ln (\frac{5}{8})

Decimal Notation

- 1.38629 \dots

Solution steps

\int_{- 3}^{3} \frac{20}{x ^{2} - 6 x - 91} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 6 x - 91} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x - 91} : - \frac{1}{20 ( x + 7 )} + \frac{1}{20 ( x - 13 )}

= 20 \cdot \int_{- 3}^{3} - \frac{1}{20 ( x + 7 )} + \frac{1}{20 ( x - 13 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 20 (- \int_{- 3}^{3} \frac{1}{20 ( x + 7 )} d x + \int_{- 3}^{3} \frac{1}{20 ( x - 13 )} d x)

\int_{- 3}^{3} \frac{1}{20 ( x + 7 )} d x = \frac{1}{20} ln (\frac{5}{2})

\int_{- 3}^{3} \frac{1}{20 ( x - 13 )} d x = \frac{1}{20} ln (\frac{5}{8})

= 20 (- \frac{1}{20} ln (\frac{5}{2}) + \frac{1}{20} ln (\frac{5}{8}))

Simplify

20 (- \frac{1}{20} ln (\frac{5}{2}) + \frac{1}{20} ln (\frac{5}{8})) : - ln (\frac{5}{2}) + ln (\frac{5}{8})

= - ln (\frac{5}{2}) + ln (\frac{5}{8})

\int \frac{x ^{2}}{( x - 3 ) ( x - 4 ) ^{2}} d x

x \to 0 lim (\frac{6 ^{x} - 8 ^{x}}{x})

y^{^{'}} = \frac{1}{y}

\frac{1}{2 \cdot 5} y + y^{^{'}} = \frac{t}{2}, y (30) = 10

(- 10, 1), (0, - 4)

What is the integral from-3 to 3 of (20)/(x^2-6x-91) ?
The integral from-3 to 3 of (20)/(x^2-6x-91) is -ln(5/2)+ln(5/8)