Question
Evaluate the following definite integrals : $\int_0^1 \frac{x^2+3 x+2}{\sqrt{x}} d x$

Answer

$
\begin{aligned}
& \int_0^1 \frac{x^2+3 x+2}{\sqrt{x}} d x \\
= & \int_0^1\left(\frac{x^2}{\sqrt{x}}+\frac{3 x}{\sqrt{x}}+\frac{2}{\sqrt{x}}\right) d x \\
= & \int_0^1\left(x^{\frac{3}{2}}+3 x^{\frac{1}{2}}+2 x^{-\frac{1}{2}}\right) d x \\
= & {\left[\frac{x^{\frac{5}{2}}}{5 / 2}+3\left(\frac{x^{\frac{3}{2}}}{3 / 2}\right)+2\left(\frac{x^{\frac{1}{2}}}{1 / 2}\right)\right]_0^1 } \\
= & {\left[\frac{2}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+4 x^{\frac{1}{2}}\right]_0^1 } \\
= & {\left[\frac{2}{5}(1)^{\frac{5}{2}}+2(1)^{\frac{3}{2}}+4(1)^{\frac{1}{2}}\right]-(0+0+0) } \\
= & \frac{2}{5}+2+4=\frac{32}{5} .
\end{aligned}
$

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