What is the integral from 1 to 64 of 1/(sqrt(x)+\sqrt[3]{x)} ?

The integral from 1 to 64 of 1/(sqrt(x)+\sqrt[3]{x)} is -6ln(3)+6ln(2)+11

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integral from 1 to 64 of 1/(sqrt(x)+\sqrt[3]{x)}

\int_{1}^{64} \frac{1}{x + 3 x} d x

- 6 ln (3) + 6 ln (2) + 11

Decimal

8.56720 \dots

Solution steps

\int_{1}^{64} \frac{1}{x + 3 x} d x

Apply u-substitution

: \int_{1}^{2} \frac{6 u ^{3}}{u + 1} d u

= \int_{1}^{2} \frac{6 u ^{3}}{u + 1} d u

Take the constant out:

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

= 6 \cdot \int_{1}^{2} \frac{u ^{3}}{u + 1} d u

Apply u-substitution

: \int_{2}^{3} \frac{( v - 1 ) ^{3}}{v} d v

= 6 \cdot \int_{2}^{3} \frac{( v - 1 ) ^{3}}{v} d v

Expand

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

= 6 \cdot \int_{2}^{3} v^{2} - 3 v + 3 - \frac{1}{v} d v

Apply the Sum Rule:

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

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

\int_{2}^{3} v^{2} d v = \frac{19}{3}

\int_{2}^{3} 3 v d v = \frac{15}{2}

\int_{2}^{3} 3 d v = 3

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

= 6 (\frac{19}{3} - \frac{15}{2} + 3 - (ln (3) - ln (2)))

Simplify

6 (\frac{19}{3} - \frac{15}{2} + 3 - (ln (3) - ln (2))) : - 6 ln (3) + 6 ln (2) + 11

= - 6 ln (3) + 6 ln (2) + 11

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What is the integral from 1 to 64 of 1/(sqrt(x)+\sqrt[3]{x)} ?
The integral from 1 to 64 of 1/(sqrt(x)+\sqrt[3]{x)} is -6ln(3)+6ln(2)+11