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

The integral from 1 to 3 of (x^2+1)/(4x-x^2) is 9/2 ln(3)-2

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

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

\frac{9}{2} ln (3) - 2

Decimal Notation

2.94375 \dots

Solution steps

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

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

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

\int_{1}^{3} \frac{4 x + 1}{4 x - x ^{2}} d x = \frac{9}{2} ln (3)

\int_{1}^{3} 1 d x = 2

= \frac{9}{2} ln (3) - 2

x y^{^{'}} - y = 3 x ln (x)

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\frac{\partial}{\partial x} (x^{2} + 2 y^{2} + 9 x z)

What is the integral from 1 to 3 of (x^2+1)/(4x-x^2) ?
The integral from 1 to 3 of (x^2+1)/(4x-x^2) is 9/2 ln(3)-2