What is the integral from 1 to 4 of (x-2)/(x(x+1)) ?

The integral from 1 to 4 of (x-2)/(x(x+1)) is -7ln(2)+3ln(5)

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integral from 1 to 4 of (x-2)/(x(x+1))

\int_{1}^{4} \frac{x - 2}{x ( x + 1 )} d x

- 7 ln (2) + 3 ln (5)

Decimal

- 0.02371 \dots

Solution steps

\int_{1}^{4} \frac{x - 2}{x ( x + 1 )} d x

Expand

\frac{x - 2}{x ( x + 1 )} : \frac{x}{x ( x + 1 )} - \frac{2}{x ( x + 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_{1}^{4} \frac{x}{x ( x + 1 )} d x - \int_{1}^{4} \frac{2}{x ( x + 1 )} d x

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

\int_{1}^{4} \frac{2}{x ( x + 1 )} d x = 2 (3 ln (2) - ln (5))

= ln (5) - ln (2) - 2 (3 ln (2) - ln (5))

Simplify

ln (5) - ln (2) - 2 (3 ln (2) - ln (5)) : - 7 ln (2) + 3 ln (5)

= - 7 ln (2) + 3 ln (5)

What is the integral from 1 to 4 of (x-2)/(x(x+1)) ?
The integral from 1 to 4 of (x-2)/(x(x+1)) is -7ln(2)+3ln(5)