What is the integral from 1 to 4 of (4+6u)/(sqrt(u)) ?

The integral from 1 to 4 of (4+6u)/(sqrt(u)) is 36

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integral from 1 to 4 of (4+6u)/(sqrt(u))

\int_{1}^{4} \frac{4 + 6 u}{u} d u

36

Solution steps

\int_{1}^{4} \frac{4 + 6 u}{u} d u

Apply Integration By Parts

: [2 u (4 + 6 u) - \int 12 u d u]_{1}^{4}

= [2 u (4 + 6 u) - \int 12 u d u]_{1}^{4}

\int 12 u d u = 8 u^{\frac{3}{2}}

= [2 u (4 + 6 u) - 8 u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

36

= 36

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What is the integral from 1 to 4 of (4+6u)/(sqrt(u)) ?
The integral from 1 to 4 of (4+6u)/(sqrt(u)) is 36