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

The integral from 1 to 4 of (3x+2)/(sqrt(x)) is 18

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

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

18

Solution steps

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

Apply Integration By Parts

: [2 x (3 x + 2) - \int 6 x d x]_{1}^{4}

= [2 x (3 x + 2) - \int 6 x d x]_{1}^{4}

\int 6 x d x = 4 x^{\frac{3}{2}}

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

Compute the boundaries:

18

= 18

\int_{0}^{2} (x^{2} - x + 1) d x

\int_{0}^{5.9} 3 3 x + 4 d x

\int_{0}^{2} (x^{2} + 5) d x

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

\int_{1}^{3} 25 - (x - 5)^{2} d x

What is the integral from 1 to 4 of (3x+2)/(sqrt(x)) ?
The integral from 1 to 4 of (3x+2)/(sqrt(x)) is 18