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

The integral from 1 to 6 of (x^2+6)/(7x-x^2) is -5+61/7 ln(6)

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

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

- 5 + \frac{61}{7} ln (6)

Decimal

10.61390 \dots

Solution steps

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

Factor

\frac{x ^{2} + 6}{7 x - x ^{2}}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{2} + 6}{x ( x - 7 )} : \frac{x ^{2}}{x ( x - 7 )} + \frac{6}{x ( x - 7 )}

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

\int_{1}^{6} \frac{x ^{2}}{x ( x - 7 )} d x = 5 - 7 ln (6)

\int_{1}^{6} \frac{6}{x ( x - 7 )} d x = - \frac{12}{7} ln (6)

= - (5 - 7 ln (6) - \frac{12}{7} ln (6))

Simplify

- (5 - 7 ln (6) - \frac{12}{7} ln (6)) : - 5 + \frac{61}{7} ln (6)

= - 5 + \frac{61}{7} ln (6)

\frac{d}{d t} (\frac{6}{5} - \frac{6}{5} e^{- 20 t})

n = 0 \sum \infty \frac{1}{n - 1}

f (x) = (e^{- x} - e^{5}) (e^{x} + e^{- 3})

n = j \sum \infty (\frac{3}{4})^{n}

\frac{1}{( s ^{2} + 16 )}

What is the integral from 1 to 6 of (x^2+6)/(7x-x^2) ?
The integral from 1 to 6 of (x^2+6)/(7x-x^2) is -5+61/7 ln(6)