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

The integral from 1 to 3 of (x^2+6)/(4x-x^2) is -2+7ln(3)

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integral from 1 to 3 of (x^2+6)/(4x-x^2)

\int_{1}^{3} \frac{x ^{2} + 6}{4 x - x ^{2}} d x

- 2 + 7 ln (3)

Decimal Notation

5.69028 \dots

Solution steps

\int_{1}^{3} \frac{x ^{2} + 6}{4 x - x ^{2}} d x

Factor

\frac{x ^{2} + 6}{4 x - x ^{2}}

= \int_{1}^{3} \frac{x ^{2} + 6}{- x ( x - 4 )} d x

Take the constant out:

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

= - \int_{1}^{3} \frac{x ^{2} + 6}{x ( x - 4 )} d x

Expand

\frac{x ^{2} + 6}{x ( x - 4 )} : \frac{x ^{2}}{x ( x - 4 )} + \frac{6}{x ( x - 4 )}

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}^{3} \frac{x ^{2}}{x ( x - 4 )} d x + \int_{1}^{3} \frac{6}{x ( x - 4 )} d x)

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

\int_{1}^{3} \frac{6}{x ( x - 4 )} d x = - 3 ln (3)

= - (2 - 4 ln (3) - 3 ln (3))

Simplify

= - 2 + 7 ln (3)

\int \frac{4 + t + t ^{2}}{t} d t

y = \frac{1}{x - 2}, at x = 6

\int e^{5} d x

\frac{d}{d x} ((2 x^{3} - 4 x^{2}) \cdot (3 x^{5} + x^{2}))

x \to i lim (\frac{z + i}{z - i})

What is the integral from 1 to 3 of (x^2+6)/(4x-x^2) ?
The integral from 1 to 3 of (x^2+6)/(4x-x^2) is -2+7ln(3)