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

The integral from 1 to 6 of (x^2+4)/(7x-x^2) is -5+57/7 ln(6)

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

\int_{1}^{6} \frac{x ^{2} + 4}{7 x - x ^{2}} d x

- 5 + \frac{57}{7} ln (6)

Decimal Notation

9.59004 \dots

Solution steps

\int_{1}^{6} \frac{x ^{2} + 4}{7 x - x ^{2}} d x

Factor

\frac{x ^{2} + 4}{7 x - x ^{2}}

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

Expand

\frac{x ^{2} + 4}{x ( x - 7 )} : \frac{x ^{2}}{x ( x - 7 )} + \frac{4}{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{4}{x ( x - 7 )} d x)

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

\int_{1}^{6} \frac{4}{x ( x - 7 )} d x = - \frac{8}{7} ln (6)

= - (5 - 7 ln (6) - \frac{8}{7} ln (6))

Simplify

- (5 - 7 ln (6) - \frac{8}{7} ln (6)) : - 5 + \frac{57}{7} ln (6)

= - 5 + \frac{57}{7} ln (6)

\frac{\partial}{\partial x} (yz ln (x))

\frac{d}{d x} ((\frac{1}{3})^{x})

\int \frac{1}{e ^{x} - 2} d x

y^{^{'}} + t y = t y^{2}

y^{^{''}} - 4 y = \frac{4 x}{e ^{2 x}}

What is the integral from 1 to 6 of (x^2+4)/(7x-x^2) ?
The integral from 1 to 6 of (x^2+4)/(7x-x^2) is -5+57/7 ln(6)