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$$\int_1^4\!\left( \frac{1}{\sqrt{x}}+\frac{1}{x}\right) \mathrm{d}x $$

The answer is $2+\ln(4)$, however I don't understand why. What I did was the following: $$\ln(x^{0.5})+\ln(x) = \ln(4^{0.5})+\ln(4).$$ I didn't use the other part of the equition, because filling in $x=1$ gives $\ln(1)$ which is zero.

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Note that $\frac{1}{\sqrt x}=x^{-\frac 12}$ and that $$\int x^\alpha dx=\frac{1}{\alpha+1}x^{\alpha+1}+C$$ for $\alpha\not=-1$.

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Hint: $$\int\! \frac{1}{\sqrt{x}}\,\mathrm{d}x=\int\!x^{-\frac{1}{2}}\, \mathrm{d}x$$

You can't just use $\ln(x^{0.5})$ as the antiderivative.

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We have $$\int_1^4 \left(\frac{1}{\sqrt{x}} + \frac{1}{x}\right) \, \mathrm{d}x = \int_1^4 \left(x^{-1/2} + \frac{1}{x}\right) \, \mathrm{d}x = \left[2\sqrt{x} + \ln x\right]_1^4 = 4 + \ln 4 - 2 = 2 + \ln 4$$

Note: You can simplify the final answer as $$2+ \ln (2^2) = 2+2\ln 2$$

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  • $\begingroup$ $\ln(4)=\ln(2^2)=2\ln(2)$ $\endgroup$ – Dr. Sonnhard Graubner Jul 5 '15 at 10:52
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    $\begingroup$ I know, but the OP has the answer as $2+ \ln 4$ and I thought it would be more appropriate to leave it in that form, lest I run the risk of confusing him/her. $\endgroup$ – Zain Patel Jul 5 '15 at 10:53
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You can't treat $\int \frac{1}{\sqrt x}dx$ simply the same as you do $\int \frac1x dx$.

I'm not going to explain the reason in detail for you here. My advice is that you carefully study the rule of variable substitution in your calculus textbook before you begin doing exercises.

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$$\int_{1}^{4}\Big (\dfrac{1}{\sqrt x}+\dfrac{1}{x} \Big )dx$$ $$=\int_{1}^{4} \dfrac{1}{\sqrt x}dx+\int_{1}^{4}\dfrac{1}{x}dx$$ $$=2\sqrt x \Big |_{1}^{4}+\ln x\Big|_{1}^{4}$$ $$=(4-2)+(\ln4-\ln1) \ \ ,\ \ \ln1=0$$ $$=2+\ln4$$

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  • $\begingroup$ You can typset the logarithmic function using \ln x in between the dollar signs. $\endgroup$ – Zain Patel Jul 5 '15 at 10:58

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