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

The integral from 1 to 5 of (x^2+3)/(6x-x^2) is -4+7ln(5)

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integral from 1 to 5 of (x^2+3)/(6x-x^2)

\int_{1}^{5} \frac{x ^{2} + 3}{6 x - x ^{2}} d x

- 4 + 7 ln (5)

Decimal

7.26606 \dots

Solution steps

\int_{1}^{5} \frac{x ^{2} + 3}{6 x - x ^{2}} d x

Factor

\frac{x ^{2} + 3}{6 x - x ^{2}}

= \int_{1}^{5} \frac{x ^{2} + 3}{- 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} + 3}{x ( x - 6 )} d x

Expand

\frac{x ^{2} + 3}{x ( x - 6 )} : \frac{x ^{2}}{x ( x - 6 )} + \frac{3}{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{3}{x ( x - 6 )} d x)

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

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

= - (4 - 6 ln (5) - ln (5))

Simplify

= - 4 + 7 ln (5)

\int t + 1 d t

\frac{d}{d x} ((3 x) (x + 3))

y^{^{''}} + 25 y = 2 sec (5 t)

x \to 0 lim (5^{x})

n = 6 \sum \infty e^{3 - 4 n}

What is the integral from 1 to 5 of (x^2+3)/(6x-x^2) ?
The integral from 1 to 5 of (x^2+3)/(6x-x^2) is -4+7ln(5)