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

The integral from 1 to 4 of (x^2+3)/(5x-x^2) is -3+62/5 ln(2)

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integral from 1 to 4 of (x^2+3)/(5x-x^2)

\int_{1}^{4} \frac{x ^{2} + 3}{5 x - x ^{2}} d x

- 3 + \frac{62}{5} ln (2)

Decimal

5.59502 \dots

Solution steps

\int_{1}^{4} \frac{x ^{2} + 3}{5 x - x ^{2}} d x

Factor

\frac{x ^{2} + 3}{5 x - x ^{2}}

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

Expand

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

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

\int_{1}^{4} \frac{3}{x ( x - 5 )} d x = - \frac{12}{5} ln (2)

= - (3 - 10 ln (2) - \frac{12}{5} ln (2))

Simplify

- (3 - 10 ln (2) - \frac{12}{5} ln (2)) : - 3 + \frac{62}{5} ln (2)

= - 3 + \frac{62}{5} ln (2)

n = 1 \sum \infty n n

f (x) = tan (\frac{3 π}{4} x)

\int a^{- x} d x

x \to 3 - lim (arcsin (\frac{x}{3}))

\frac{1}{3 x}

What is the integral from 1 to 4 of (x^2+3)/(5x-x^2) ?
The integral from 1 to 4 of (x^2+3)/(5x-x^2) is -3+62/5 ln(2)