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

The integral from 1 to 4 of (x^2+6)/(5x-x^2) is -3+74/5 ln(2)

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

\int_{1}^{4} \frac{x ^{2} + 6}{5 x - x ^{2}} d x

- 3 + \frac{74}{5} ln (2)

Decimal Notation

7.25857 \dots

Solution steps

\int_{1}^{4} \frac{x ^{2} + 6}{5 x - x ^{2}} d x

Factor

\frac{x ^{2} + 6}{5 x - x ^{2}}

= \int_{1}^{4} \frac{x ^{2} + 6}{- x ( x - 5 )} d x

Take the constant out:

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

= - \int_{1}^{4} \frac{x ^{2} + 6}{x ( x - 5 )} d x

Expand

\frac{x ^{2} + 6}{x ( x - 5 )} : \frac{x ^{2}}{x ( x - 5 )} + \frac{6}{x ( x - 5 )}

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}^{4} \frac{x ^{2}}{x ( x - 5 )} d x + \int_{1}^{4} \frac{6}{x ( x - 5 )} d x)

\int_{1}^{4} \frac{x ^{2}}{x ( x - 5 )} d x = 3 - 10 ln (2)

\int_{1}^{4} \frac{6}{x ( x - 5 )} d x = - \frac{24}{5} ln (2)

= - (3 - 10 ln (2) - \frac{24}{5} ln (2))

Simplify

- (3 - 10 ln (2) - \frac{24}{5} ln (2)) : - 3 + \frac{74}{5} ln (2)

= - 3 + \frac{74}{5} ln (2)

\int cos^{2} (π) x d x

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

\int x^{7} sin (1 + x^{\frac{9}{2}}) d x

x \to 3 lim (\frac{7}{( x - 3 ) ^{2}})

f (x) = \frac{x ^{2}}{x + 6}

What is the integral from 1 to 4 of (x^2+6)/(5x-x^2) ?
The integral from 1 to 4 of (x^2+6)/(5x-x^2) is -3+74/5 ln(2)