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

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

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

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

6 ln (5) + \frac{2 ln ( 5 )}{3} - 4

Decimal

6.72958 \dots

Solution steps

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

\frac{x ^{2} + 2}{6 x - x ^{2}} = \frac{6 x + 2}{6 x - x ^{2}} - 1

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

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

\int_{1}^{5} 1 d x = 4

= 6 ln (5) + \frac{2 ln ( 5 )}{3} - 4

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