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

The integral from 2 to 6 of (x-4)/(x+1) is 5ln(3)-5ln(7)+4

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integral from 2 to 6 of (x-4)/(x+1)

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

5 ln (3) - 5 ln (7) + 4

Solution steps

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

Expand

\frac{x - 4}{x + 1} : \frac{x}{x + 1} - \frac{4}{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_{2}^{6} \frac{x}{x + 1} d x - \int_{2}^{6} \frac{4}{x + 1} d x

\int_{2}^{6} \frac{x}{x + 1} d x = 4 - ln (7) + ln (3)

\int_{2}^{6} \frac{4}{x + 1} d x = 4 (ln (7) - ln (3))

= 4 - ln (7) + ln (3) - 4 (ln (7) - ln (3))

Simplify

4 - ln (7) + ln (3) - 4 (ln (7) - ln (3)) : 5 ln (3) - 5 ln (7) + 4

= 5 ln (3) - 5 ln (7) + 4

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\frac{d y}{3 x + e ^{x}} = e^{- y} d x

csc (x) (x + cot (x))

What is the integral from 2 to 6 of (x-4)/(x+1) ?
The integral from 2 to 6 of (x-4)/(x+1) is 5ln(3)-5ln(7)+4