What is the integral from 1 to 2 of (7x-2)/(3x^2-2x) ?

The integral from 1 to 2 of (7x-2)/(3x^2-2x) is 11/3 ln(2)

integral from 1 to 2 of (7x-2)/(3x^2-2x)

\int_{1}^{2} \frac{7 x - 2}{3 x ^{2} - 2 x} d x

\frac{11}{3} ln (2)

Decimal

2.54153 \dots

Solution steps

\int_{1}^{2} \frac{7 x - 2}{3 x ^{2} - 2 x} d x

Expand

\frac{7 x - 2}{3 x ^{2} - 2 x} : \frac{7 x}{3 x ^{2} - 2 x} - \frac{2}{3 x ^{2} - 2 x}

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}^{2} \frac{7 x}{3 x ^{2} - 2 x} d x - \int_{1}^{2} \frac{2}{3 x ^{2} - 2 x} d x

\int_{1}^{2} \frac{7 x}{3 x ^{2} - 2 x} d x = \frac{14}{3} ln (2)

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

= \frac{14}{3} ln (2) - ln (2)

Simplify

\frac{14}{3} ln (2) - ln (2) : \frac{11}{3} ln (2)

= \frac{11}{3} ln (2)

t an g e n t f (x) = (4 x + 3) (- x + 1), at x = 4

\frac{d}{d x} (\frac{6}{4 x})

t an g e n t f (x) = 7 x - x^{2}, (1, 6)

\int_{- 1}^{1} \frac{1}{x} d x

a re a y = \frac{x ^{2} + 5}{6 x - x ^{2}}, (1, 5)

What is the integral from 1 to 2 of (7x-2)/(3x^2-2x) ?
The integral from 1 to 2 of (7x-2)/(3x^2-2x) is 11/3 ln(2)