What is the integral from 0 to 5 of (3x-4)/(x+1) ?

The integral from 0 to 5 of (3x-4)/(x+1) is 15-7ln(6)

Popular >

integral from 0 to 5 of (3x-4)/(x+1)

\int_{0}^{5} \frac{3 x - 4}{x + 1} d x

15 - 7 ln (6)

Decimal Notation

2.45768 \dots

Solution steps

\int_{0}^{5} \frac{3 x - 4}{x + 1} d x

Apply u-substitution

: \int_{1}^{6} \frac{3 u - 7}{u} d u

= \int_{1}^{6} \frac{3 u - 7}{u} d u

Expand

\frac{3 u - 7}{u} : 3 - \frac{7}{u}

= \int_{1}^{6} 3 - \frac{7}{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}^{6} 3 d u - \int_{1}^{6} \frac{7}{u} d u

\int_{1}^{6} 3 d u = 15

\int_{1}^{6} \frac{7}{u} d u = 7 ln (6)

= 15 - 7 ln (6)

x \to 0 - lim (\frac{1}{x ^{2}} - \frac{1}{x})

\int_{1}^{3} e^{x^{2}} d x

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

\int 2 e^{- 4 x} d x

\int \frac{1 - x}{x} d x

What is the integral from 0 to 5 of (3x-4)/(x+1) ?
The integral from 0 to 5 of (3x-4)/(x+1) is 15-7ln(6)