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

The integral from 1 to 2 of (x^2+2)/(3x-x^2) is 3ln(2)+(4ln(2))/3-1

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

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

3 ln (2) + \frac{4 ln ( 2 )}{3} - 1

Decimal

2.00363 \dots

Solution steps

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

\frac{x ^{2} + 2}{3 x - x ^{2}} = \frac{3 x + 2}{3 x - x ^{2}} - 1

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

\int_{1}^{2} \frac{3 x + 2}{3 x - x ^{2}} d x = 3 ln (2) + \frac{4 ln ( 2 )}{3}

\int_{1}^{2} 1 d x = 1

= 3 ln (2) + \frac{4 ln ( 2 )}{3} - 1

What is the integral from 1 to 2 of (x^2+2)/(3x-x^2) ?
The integral from 1 to 2 of (x^2+2)/(3x-x^2) is 3ln(2)+(4ln(2))/3-1