Find the area under the curve of $y\,=\,{\frac {8}{\sqrt {x}}}$ between $x=1$ and $x=4$.
Foundations:

This problem requires two rules of integration. In particular, you need

Integration by substitution (U  sub): If $f$ and $g$ are differentiable functions, then

$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$

The Product Rule: If $f$ and $g$ are differentiable functions, then

$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$

The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then

$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$

Additionally, we will need our power rule for differentiation:

 $\left(x^{n}\right)'\,=\,nx^{n1},$ for $n\neq 0$,

as well as the derivative of natural log:

 $\left(\ln x\right)'\,=\,{\frac {1}{x}}.$


Solution:
Step 1:

Set up the integral:

$\int _{1}^{4}{\frac {8}{\sqrt {x}}}dx$

Step 2:

Using the power rule we have:

 Failed to parse (unknown function "\begin{array}"): {\displaystyle \begin{array}{rcl} \int_1^4 \frac{8}{\sqrt{x}}dx & = & \frac{x^{1/2}} \end{array}}

Step 3:

Now we need to substitute back into our original variables using our original substitution $u=3x+2$

to get ${\frac {\log(u)}{3}}={\frac {\log(3x+2}{3}}$

Step 4:

Since this integral is an indefinite integral we have to remember to add "+ C" at the end.

Final Answer:

$\int {\frac {1}{3x+2}}dx={\frac {\ln(3x+2)}{3}}+C$

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