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

The integral from 2 to 3 of (3x^3)/(x-1) is 3(ln(2)+59/6)

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

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

3 (ln (2) + \frac{59}{6})

Decimal

31.57944 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 3 \cdot \int_{1}^{2} \frac{( u + 1 ) ^{3}}{u} d u

Expand

\frac{( u + 1 ) ^{3}}{u} : u^{2} + 3 u + 3 + \frac{1}{u}

= 3 \cdot \int_{1}^{2} u^{2} + 3 u + 3 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int_{1}^{2} u^{2} d u + \int_{1}^{2} 3 u d u + \int_{1}^{2} 3 d u + \int_{1}^{2} \frac{1}{u} d u)

\int_{1}^{2} u^{2} d u = \frac{7}{3}

\int_{1}^{2} 3 u d u = \frac{9}{2}

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

\int_{1}^{2} \frac{1}{u} d u = ln (2)

= 3 (\frac{7}{3} + \frac{9}{2} + 3 + ln (2))

Combine the fractions

3 + \frac{7}{3} + \frac{9}{2} : \frac{59}{6}

= 3 (ln (2) + \frac{59}{6})

What is the integral from 2 to 3 of (3x^3)/(x-1) ?
The integral from 2 to 3 of (3x^3)/(x-1) is 3(ln(2)+59/6)