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

The integral from 1 to 4 of (x^2+1)/(5x-x^2) is 10ln(2)+(4ln(2))/5-3

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

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

10 ln (2) + \frac{4 ln ( 2 )}{5} - 3

Decimal Notation

4.48598 \dots

Solution steps

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

\frac{x ^{2} + 1}{5 x - x ^{2}} = \frac{5 x + 1}{5 x - x ^{2}} - 1

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

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

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

\int_{1}^{4} 1 d x = 3

= 10 ln (2) + \frac{4 ln ( 2 )}{5} - 3

\frac{d y}{d x}, \frac{( x ^{2} )}{4} + y^{2} = 1

\frac{d}{d x} (\frac{1}{4 x ^{3}})

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

\int_{- 4}^{2} (12 - x - x^{2}) - (x + 4) d x

\frac{\partial}{\partial x} (x^{5} + 7 x^{5} y)

What is the integral from 1 to 4 of (x^2+1)/(5x-x^2) ?
The integral from 1 to 4 of (x^2+1)/(5x-x^2) is 10ln(2)+(4ln(2))/5-3