What is the integral from 1 to 5 of (x^2+4)/(6x-x^2) ?

The integral from 1 to 5 of (x^2+4)/(6x-x^2) is -4+22/3 ln(5)

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integral from 1 to 5 of (x^2+4)/(6x-x^2)

\int_{1}^{5} \frac{x ^{2} + 4}{6 x - x ^{2}} d x

- 4 + \frac{22}{3} ln (5)

Decimal Notation

7.80254 \dots

Solution steps

\int_{1}^{5} \frac{x ^{2} + 4}{6 x - x ^{2}} d x

Factor

\frac{x ^{2} + 4}{6 x - x ^{2}}

= \int_{1}^{5} \frac{x ^{2} + 4}{- x ( x - 6 )} d x

Take the constant out:

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

= - \int_{1}^{5} \frac{x ^{2} + 4}{x ( x - 6 )} d x

Expand

\frac{x ^{2} + 4}{x ( x - 6 )} : \frac{x ^{2}}{x ( x - 6 )} + \frac{4}{x ( x - 6 )}

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}^{5} \frac{x ^{2}}{x ( x - 6 )} d x + \int_{1}^{5} \frac{4}{x ( x - 6 )} d x)

\int_{1}^{5} \frac{x ^{2}}{x ( x - 6 )} d x = 4 - 6 ln (5)

\int_{1}^{5} \frac{4}{x ( x - 6 )} d x = - \frac{4}{3} ln (5)

= - (4 - 6 ln (5) - \frac{4}{3} ln (5))

Simplify

- (4 - 6 ln (5) - \frac{4}{3} ln (5)) : - 4 + \frac{22}{3} ln (5)

= - 4 + \frac{22}{3} ln (5)

f (x) = x^{2} - 1, at x = 4

y = 2 x^{3} - 5 x, at x = (3 - 2)

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

y^{^{'}} = \frac{( 2 - y ) ^{2}}{2 \cdot 1 + x}

x = 2 y^{2}, x = 4 + y^{2}

What is the integral from 1 to 5 of (x^2+4)/(6x-x^2) ?
The integral from 1 to 5 of (x^2+4)/(6x-x^2) is -4+22/3 ln(5)