What is the integral from 3 to 4 of (x+4)/(x^2(x-2)) ?

The integral from 3 to 4 of (x+4)/(x^2(x-2)) is -(ln(2)-ln(3))/2-1/6-ln(2)+ln(3)

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integral from 3 to 4 of (x+4)/(x^2(x-2))

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

- \frac{ln ( 2 ) - ln ( 3 )}{2} - \frac{1}{6} - ln (2) + ln (3)

Decimal

0.44153 \dots

Solution steps

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

Expand

\frac{x + 4}{x ^{2} ( x - 2 )} : \frac{x}{x ^{2} ( x - 2 )} + \frac{4}{x ^{2} ( x - 2 )}

Apply the Sum Rule:

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

= \int_{3}^{4} \frac{x}{x ^{2} ( x - 2 )} d x + \int_{3}^{4} \frac{4}{x ^{2} ( x - 2 )} d x

\int_{3}^{4} \frac{x}{x ^{2} ( x - 2 )} d x = - \frac{ln ( 2 ) - ln ( 3 )}{2}

\int_{3}^{4} \frac{4}{x ^{2} ( x - 2 )} d x = - \frac{1}{6} - ln (2) + ln (3)

= - \frac{ln ( 2 ) - ln ( 3 )}{2} - \frac{1}{6} - ln (2) + ln (3)

\int_{- 2}^{4} x^{2} - 2 x - 3 d x

\int_{0}^{\frac{π}{8}} tan (4 x) d x

\int_{1}^{2} \frac{x ^{2}}{6 x ^{3} + 1} d x

\int_{0}^{\frac{π}{2}} (x + 1) cos (x) d x

\int_{0}^{7} \frac{1}{x - 2} d x

What is the integral from 3 to 4 of (x+4)/(x^2(x-2)) ?
The integral from 3 to 4 of (x+4)/(x^2(x-2)) is -(ln(2)-ln(3))/2-1/6-ln(2)+ln(3)