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

The integral from 1 to 5 of (x^2+1)/(6x-x^2) is 19/3 ln(5)-4

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

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

\frac{19}{3} ln (5) - 4

Decimal Notation

6.19310 \dots

Solution steps

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

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

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

\int_{1}^{5} \frac{6 x + 1}{6 x - x ^{2}} d x = \frac{19}{3} ln (5)

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

= \frac{19}{3} ln (5) - 4

\frac{d}{d x} (6 x^{2} - 8 + 40 cos (π) x)

\int \frac{x ^{3}}{1 + x ^{2}} d x

\frac{d}{d x} (\frac{2 x + 1}{x})

f (x) = \frac{( x + 1 )}{x - 1}

\frac{\partial}{\partial y} (x + 1)

What is the integral from 1 to 5 of (x^2+1)/(6x-x^2) ?
The integral from 1 to 5 of (x^2+1)/(6x-x^2) is 19/3 ln(5)-4