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

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

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

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

3 ln (2) + \frac{2 ln ( 2 )}{3} - 1

Decimal Notation

1.54153 \dots

Solution steps

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

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

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

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

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

= 3 ln (2) + \frac{2 ln ( 2 )}{3} - 1

\frac{d}{d x} ((x + (x + sin^{2} (x))^{3})^{4})

x \to - 4 lim (\frac{x ^{2}}{4 x + 16})

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

\int sin (x^{\frac{1}{2}}) d x

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