What is the integral from 0 to 3 of (2x-1)/(x+1) ?

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

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integral from 0 to 3 of (2x-1)/(x+1)

\int_{0}^{3} \frac{2 x - 1}{x + 1} d x

6 - 6 ln (2)

Decimal Notation

1.84111 \dots

Solution steps

\int_{0}^{3} \frac{2 x - 1}{x + 1} d x

Apply u-substitution

: \int_{1}^{4} \frac{2 u - 3}{u} d u

= \int_{1}^{4} \frac{2 u - 3}{u} d u

Expand

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

= \int_{1}^{4} 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}^{4} 2 d u - \int_{1}^{4} \frac{3}{u} d u

\int_{1}^{4} 2 d u = 6

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

= 6 - 6 ln (2)

\frac{d}{d x} (\frac{42}{x})

\frac{d}{d x} (ln (\frac{( e ^{x} + 1 )}{e ^{x} - 1}))

ln (x^{2} + 2)

y^{^{'}} = 1 + 2 x + y^{2} + 2 x y^{2}

\frac{1}{s - 10}

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