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

The integral from 1 to 2 of (x^2+4)/(3x-x^2) is -1+17/3 ln(2)

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

\int_{1}^{2} \frac{x ^{2} + 4}{3 x - x ^{2}} d x

- 1 + \frac{17}{3} ln (2)

Decimal Notation

2.92783 \dots

Solution steps

\int_{1}^{2} \frac{x ^{2} + 4}{3 x - x ^{2}} d x

Factor

\frac{x ^{2} + 4}{3 x - x ^{2}}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{2} + 4}{x ( x - 3 )} : \frac{x ^{2}}{x ( x - 3 )} + \frac{4}{x ( x - 3 )}

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

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

\int_{1}^{2} \frac{4}{x ( x - 3 )} d x = - \frac{8}{3} ln (2)

= - (1 - 3 ln (2) - \frac{8}{3} ln (2))

Simplify

- (1 - 3 ln (2) - \frac{8}{3} ln (2)) : - 1 + \frac{17}{3} ln (2)

= - 1 + \frac{17}{3} ln (2)

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What is the integral from 1 to 2 of (x^2+4)/(3x-x^2) ?
The integral from 1 to 2 of (x^2+4)/(3x-x^2) is -1+17/3 ln(2)