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

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

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

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

7 ln (6) + \frac{4 ln ( 6 )}{7} - 5

Decimal Notation

8.56617 \dots

Solution steps

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

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

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

\int_{1}^{6} \frac{7 x + 2}{7 x - x ^{2}} d x = 7 ln (6) + \frac{4 ln ( 6 )}{7}

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

= 7 ln (6) + \frac{4 ln ( 6 )}{7} - 5

\int \frac{1}{sec ( θ )} d θ

x + x (y^{^{'}}) 2 - 1 = 0

y^{^{''}} + 2 y^{^{'}} = 3

\frac{d}{d x} (y - 3 x)

\frac{d}{d x} (2 x^{3} + \frac{1}{x ^{2}} + 16 x^{\frac{7}{2}} + 5)

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