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Helga [31]
3 years ago
9

n^{p}x }" alt=" \int \frac{dx}{xln^{p}x }" align="absmiddle" class="latex-formula"> How would i integrate this? I tried integration by parts but I couldn't figure out a useful way to use it
Mathematics
1 answer:
Eva8 [605]3 years ago
4 0
Substitute u=\ln x, so that \mathrm du=\dfrac{\mathrm dx}x. The integral is then equivalent to

\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\int\frac{\mathrm du}{u^p}=\begin{cases}\dfrac{u^{p+1}}{p+1}+C&\text{for }p\neq1\\\\\ln|u|+C&\text{for }p=1\end{cases}

Then transforming back to x gives

\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\begin{cases}\dfrac{\ln^{p+1}x}{p+1}+C&\text{for }p\neq1\\\\\ln|\ln x|+C&\text{for }p=1\end{cases}
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Lisa wants to use her calculator to square a two-digit positive integer, but she accidentally enters the tens digit incorrectly.
Svetradugi [14.3K]

Answer: Correct two digit number= 24

Step-by-step explanation:

Let

x= tens digit (wrong one)

and y= zeroes digit

and z= tens digit (right one)

Now according to the statement,

(10x+y)² -  (10z+y)² = 2340

Simplifying this we get

(x-z)(5x+y+5z)= 117= 1 * 3 * 3 * 13

So it could either be

x-z=1 or x-z=3

If we take the x-z=1

then 5x+y+5z=117

which is not possible as maximum value we can get from this is 99.

This leaves us with x-z=3

so 5x+y+5z= 39

Put x=3+z

We get,

y+10z= 24

Now as y and z both are single digits so the only possible numbers to fit these are y=4 and z=2.

Put z=2 in x=3+z , we get x=5

So the numbers become 10x+y= 54

and 10z+y= 24

Their sum is 54+24=78

Check

54²-24²= 2340 so it is correct.

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mrs_skeptik [129]

Answer:

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An irrational number is any real number that cannot be expressed as a simple fraction or terminating decimals. There are many answers to this question.
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Step-by-step explanation:

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