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

The integral from 1 to 4 of (x^2+7)/(5x-x^2) is -3+78/5 ln(2)

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

\int_{1}^{4} \frac{x ^{2} + 7}{5 x - x ^{2}} d x

- 3 + \frac{78}{5} ln (2)

Decimal

7.81309 \dots

Solution steps

\int_{1}^{4} \frac{x ^{2} + 7}{5 x - x ^{2}} d x

Factor

\frac{x ^{2} + 7}{5 x - x ^{2}}

= \int_{1}^{4} \frac{x ^{2} + 7}{- 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} + 7}{x ( x - 5 )} d x

Expand

\frac{x ^{2} + 7}{x ( x - 5 )} : \frac{x ^{2}}{x ( x - 5 )} + \frac{7}{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{7}{x ( x - 5 )} d x)

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

\int_{1}^{4} \frac{7}{x ( x - 5 )} d x = - \frac{28}{5} ln (2)

= - (3 - 10 ln (2) - \frac{28}{5} ln (2))

Simplify

- (3 - 10 ln (2) - \frac{28}{5} ln (2)) : - 3 + \frac{78}{5} ln (2)

= - 3 + \frac{78}{5} ln (2)

y^{^{'}} = e^{- y} (x^{2} - 6)

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 3

\int e^{x} 36 - e^{2 x} d x

\frac{4 e ^{x}}{cos ( x )}

y^{^{''}} - 4 y^{^{'}} + 13 y = 2 e^{x}

What is the integral from 1 to 4 of (x^2+7)/(5x-x^2) ?
The integral from 1 to 4 of (x^2+7)/(5x-x^2) is -3+78/5 ln(2)