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

The integral from 1 to 2 of (x^2+7)/(3x-x^2) is -1+23/3 ln(2)

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

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

- 1 + \frac{23}{3} ln (2)

Decimal Notation

4.31412 \dots

Solution steps

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

Factor

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

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

Expand

\frac{x ^{2} + 7}{x ( x - 3 )} : \frac{x ^{2}}{x ( x - 3 )} + \frac{7}{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{7}{x ( x - 3 )} d x)

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

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

= - (1 - 3 ln (2) - \frac{14}{3} ln (2))

Simplify

- (1 - 3 ln (2) - \frac{14}{3} ln (2)) : - 1 + \frac{23}{3} ln (2)

= - 1 + \frac{23}{3} ln (2)

\frac{s}{( 2 s + 1 ) ( ( s + 2 ) ^{2} - 1 )}

f (x) = x^{3} - x, (- 1, 0)

\int \frac{1}{3 x + 8} d x

\int x^{2} + 16 d x

g (x) = (\frac{3 x ^{2} - 2}{2 x + 3})^{3}

What is the integral from 1 to 2 of (x^2+7)/(3x-x^2) ?
The integral from 1 to 2 of (x^2+7)/(3x-x^2) is -1+23/3 ln(2)