∫94(x+1x)2dx

Question

∫94(x+1x)2dx

svartmaleJ · Accepted Answer

Compute the definite integral:
For the integrand ${(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2}$ , do long division:
$= \int_{4}^{9} x + \frac{1}{x} + 2 d x$
Integrate the sum term by term and factor out constants:
$= \int_{4}^{9} x d x + \int_{4}^{9} \frac{1}{x} d x + 2 \int_{4}^{9} 1 d x$
Apply the fundamental theorem of calculus.
The antiderivative of x is $\frac{x^{2}}{2}$ :
$= \frac{x^{2}}{2}$ right bracketing bar $_{4}^{9} + \int_{4}^{9} \frac{1}{x} d x + 2 \int_{4}^{9} 1 d x$
Evaluate the antiderivative at the limits and subtract.
$\frac{x^{2}}{2}$ right bracketing bar $_{4}^{9} = \frac{9^{2}}{2} - \frac{4^{2}}{2} = \frac{65}{2} :$
$= \frac{65}{2} + \int_{4}^{9} \frac{1}{x} d x + 2 \int_{4}^{9} 1 d x$
Apply the fundamental theorem of calculus.
The antiderivative of $\frac{1}{x}$ is $\log (x)$ :
$= \frac{65}{2} + \log (x)$ right bracketing bar $_{4}^{9} + 2 \int_{4}^{9} 1 d x$
Evaluate the antiderivative at the limits and subtract.
$\log (x)$ right bracketing bar $_{4}^{9} = \log (9) - \log (4) = \log (\frac{9}{4})$ :
$= \frac{65}{2} + \log (\frac{9}{4}) + 2 \int_{4}^{9} 1 d x$
Apply the fundamental theorem of calculus.
The antiderivative of 1 is x:
$= \frac{65}{2} + \log (\frac{9}{4}) + 2 x$ right bracketing bar $_{4}^{9}$
Evaluate the antiderivative at the limits and subtract.
2x right bracketing bar $_{4}^{9} = 29 - 24 = 10$ :
Answer:
$= \frac{85}{2} + \log (\frac{9}{4})$

int_{9}^{4} (sqrt x+frac{1}{sqrt x})^{2}dx

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