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

The integral from 0 to 3 of (3x-1)/(x+1) is 9-8ln(2)

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

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

9 - 8 ln (2)

Decimal Notation

3.45482 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{3 u - 4}{u} : 3 - \frac{4}{u}

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

\int_{1}^{4} 3 d u = 9

\int_{1}^{4} \frac{4}{u} d u = 8 ln (2)

= 9 - 8 ln (2)

\int_{- 3}^{1} 3 x^{2} + x - 10 d x

\int_{- 2}^{3} arctan (x) d x

\int_{0}^{3} 9 e^{- 2 x} d x

\int_{x}^{y} 2^{l n (t)} d t

\int_{- 2}^{2} x^{2} + a x^{3} + b x^{4} d x

What is the integral from 0 to 3 of (3x-1)/(x+1) ?
The integral from 0 to 3 of (3x-1)/(x+1) is 9-8ln(2)