What is the integral from 0 to 1 of (x-4)/(x^2-5x-6) ?

The integral from 0 to 1 of (x-4)/(x^2-5x-6) is (5ln(2)+2(ln(5)-ln(6)))/7

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integral from 0 to 1 of (x-4)/(x^2-5x-6)

\int_{0}^{1} \frac{x - 4}{x ^{2} - 5 x - 6} d x

\frac{5 ln ( 2 ) + 2 ( ln ( 5 ) - ln ( 6 ) )}{7}

Decimal Notation

0.44301 \dots

Solution steps

\int_{0}^{1} \frac{x - 4}{x ^{2} - 5 x - 6} d x

Take the partial fraction of

\frac{x - 4}{x ^{2} - 5 x - 6} : \frac{5}{7 ( x + 1 )} + \frac{2}{7 ( x - 6 )}

= \int_{0}^{1} \frac{5}{7 ( x + 1 )} + \frac{2}{7 ( x - 6 )} 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

= \int_{0}^{1} \frac{5}{7 ( x + 1 )} d x + \int_{0}^{1} \frac{2}{7 ( x - 6 )} d x

\int_{0}^{1} \frac{5}{7 ( x + 1 )} d x = \frac{5}{7} ln (2)

\int_{0}^{1} \frac{2}{7 ( x - 6 )} d x = \frac{2}{7} (ln (5) - ln (6))

= \frac{5}{7} ln (2) + \frac{2}{7} (ln (5) - ln (6))

Simplify

\frac{5}{7} ln (2) + \frac{2}{7} (ln (5) - ln (6)) : \frac{5 ln ( 2 ) + 2 ( ln ( 5 ) - ln ( 6 ) )}{7}

= \frac{5 ln ( 2 ) + 2 ( ln ( 5 ) - ln ( 6 ) )}{7}

\int \frac{1}{( x ^{2} + 1 ) ^{3}} d x

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

\frac{2000}{( 2000 - P ) P} P^{^{'}} = 0.05

30

y^{^{''}} + y^{^{'}} - 5 x = 0

What is the integral from 0 to 1 of (x-4)/(x^2-5x-6) ?
The integral from 0 to 1 of (x-4)/(x^2-5x-6) is (5ln(2)+2(ln(5)-ln(6)))/7