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

The integral from 1 to 3 of (x^2+7)/(4x-x^2) is -2+15/2 ln(3)

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

\int_{1}^{3} \frac{x ^{2} + 7}{4 x - x ^{2}} d x

- 2 + \frac{15}{2} ln (3)

Decimal Notation

6.23959 \dots

Solution steps

\int_{1}^{3} \frac{x ^{2} + 7}{4 x - x ^{2}} d x

Factor

\frac{x ^{2} + 7}{4 x - x ^{2}}

= \int_{1}^{3} \frac{x ^{2} + 7}{- 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} + 7}{x ( x - 4 )} d x

Expand

\frac{x ^{2} + 7}{x ( x - 4 )} : \frac{x ^{2}}{x ( x - 4 )} + \frac{7}{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{7}{x ( x - 4 )} d x)

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

\int_{1}^{3} \frac{7}{x ( x - 4 )} d x = - \frac{7}{2} ln (3)

= - (2 - 4 ln (3) - \frac{7}{2} ln (3))

Simplify

- (2 - 4 ln (3) - \frac{7}{2} ln (3)) : - 2 + \frac{15}{2} ln (3)

= - 2 + \frac{15}{2} ln (3)

\frac{\partial}{\partial x} (2 x y e^{- x^{2} - y^{2}})

\frac{d}{d x} (6 sec (x) tan (x))

\int_{0}^{\infty} e^{\frac{x}{3}} d x

x^{2} \frac{d y}{d x} + 4 x y - y^{3} = 0

\int_{0}^{1} 2 π (x) (8 - 7 x - x^{2}) d x

What is the integral from 1 to 3 of (x^2+7)/(4x-x^2) ?
The integral from 1 to 3 of (x^2+7)/(4x-x^2) is -2+15/2 ln(3)