What is the integral from 1 to 6 of (x^2+1)/(7x-x^2) ?

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

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integral from 1 to 6 of (x^2+1)/(7x-x^2)

\int_{1}^{6} \frac{x ^{2} + 1}{7 x - x ^{2}} d x

7 ln (6) + \frac{2 ln ( 6 )}{7} - 5

Decimal

8.05424 \dots

Solution steps

\int_{1}^{6} \frac{x ^{2} + 1}{7 x - x ^{2}} d x

\frac{x ^{2} + 1}{7 x - x ^{2}} = \frac{7 x + 1}{7 x - x ^{2}} - 1

= \int_{1}^{6} \frac{7 x + 1}{7 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}^{6} \frac{7 x + 1}{7 x - x ^{2}} d x - \int_{1}^{6} 1 d x

\int_{1}^{6} \frac{7 x + 1}{7 x - x ^{2}} d x = 7 ln (6) + \frac{2 ln ( 6 )}{7}

\int_{1}^{6} 1 d x = 5

= 7 ln (6) + \frac{2 ln ( 6 )}{7} - 5

\frac{d}{d x} (π x)

\frac{d y}{d x} + \frac{4}{x} y = - 3 x

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

\frac{d}{d x} (\frac{e ^{x}}{( 1 - e ^{x} ) ^{2}})

f (t) = - 2

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