What is the integral from 0 to 8 of (2x-1)/(x+1) ?

The integral from 0 to 8 of (2x-1)/(x+1) is 16-6ln(3)

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integral from 0 to 8 of (2x-1)/(x+1)

\int_{0}^{8} \frac{2 x - 1}{x + 1} d x

16 - 6 ln (3)

Decimal

9.40832 \dots

Solution steps

\int_{0}^{8} \frac{2 x - 1}{x + 1} d x

Apply u-substitution

: \int_{1}^{9} \frac{2 u - 3}{u} d u

= \int_{1}^{9} \frac{2 u - 3}{u} d u

Expand

\frac{2 u - 3}{u} : 2 - \frac{3}{u}

= \int_{1}^{9} 2 - \frac{3}{u} d u

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}^{9} 2 d u - \int_{1}^{9} \frac{3}{u} d u

\int_{1}^{9} 2 d u = 16

\int_{1}^{9} \frac{3}{u} d u = 6 ln (3)

= 16 - 6 ln (3)

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

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

\frac{d}{d x} (2 e^{- 2 x} x - 2 e^{- 2 x} x^{2})

What is the integral from 0 to 8 of (2x-1)/(x+1) ?
The integral from 0 to 8 of (2x-1)/(x+1) is 16-6ln(3)