What is the integral of (x^2-5x+2)/(3sqrt(x)) ?

The integral of (x^2-5x+2)/(3sqrt(x)) is 1/3 (2/5 x^{5/2}-10/3 x^{3/2}+4sqrt(x))+C

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integral of (x^2-5x+2)/(3sqrt(x))

\int \frac{x ^{2} - 5 x + 2}{3 x} d x

\frac{1}{3} (\frac{2}{5} x^{\frac{5}{2}} - \frac{10}{3} x^{\frac{3}{2}} + 4 x) + C

Solution steps

\int \frac{x ^{2} - 5 x + 2}{3 x} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{x ^{2} - 5 x + 2}{x} d x

Expand

\frac{x ^{2} - 5 x + 2}{x} : x^{\frac{3}{2}} - 5 x + \frac{2}{x}

= \frac{1}{3} \cdot \int x^{\frac{3}{2}} - 5 x + \frac{2}{x} 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

= \frac{1}{3} (\int x^{\frac{3}{2}} d x - \int 5 x d x + \int \frac{2}{x} d x)

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

\int 5 x d x = \frac{10}{3} x^{\frac{3}{2}}

\int \frac{2}{x} d x = 4 x

= \frac{1}{3} (\frac{2}{5} x^{\frac{5}{2}} - \frac{10}{3} x^{\frac{3}{2}} + 4 x)

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} x^{\frac{5}{2}} - \frac{10}{3} x^{\frac{3}{2}} + 4 x) + C

f (x) = - 2 x^{3} + x^{2}, at x = 1

y = 10 x, (10, 10)

\frac{d}{d x} (ln (5 - x^{2}))

\int_{2}^{4} \frac{1}{16 - t ^{2}} d t

x \to 2 lim (2 x - 2)

What is the integral of (x^2-5x+2)/(3sqrt(x)) ?
The integral of (x^2-5x+2)/(3sqrt(x)) is 1/3 (2/5 x^{5/2}-10/3 x^{3/2}+4sqrt(x))+C