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- Thread starter pierce15
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Infrared

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[itex] \int {e^{x+e^x}}\,dx [/itex]

[itex] \int {\frac{dx}{1- e^x}}\ [/itex]

[itex] \int {\frac{e^{2x}}{1+ e^x}}\,dx [/itex]

[itex] \int {\frac{x^2}{x^2+1}}\,dx [/itex]

[itex] \int {\frac{1}{1+ x^4}}\,dx [/itex] <------- warning, really tedious

I don't know of any good series problems, sorry.

- #3

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[itex] \int {e^{x+e^x}}\,dx [/itex]

[itex] \int {\frac{dx}{1- e^x}}\ [/itex]

[itex] \int {\frac{e^{2x}}{1+ e^x}}\,dx [/itex]

[itex] \int {\frac{x^2}{x^2+1}}\,dx [/itex]

[itex] \int {\frac{1}{1+ x^4}}\,dx [/itex] <------- warning, really tedious

I don't know of any good series problems, sorry.

For the first one, I did some manipulations and made the substitution u=e^e^x, which made it just an indefinite integral of du. Thus, my answer is e^e^x + C

For the fourth one I did a partial frac decomp and a trig substitution. Final answer:

[tex] x - \arctan(x) + C [/tex]

For some reason, I don't see anything that I can do with the second and third ones.

The fourth one is something that I'll work on in the next hour or so, I'll post an answer if I get one.

- #4

Infrared

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There was a very slick way of doing the fourth one that I don't blame you for not getting.

[itex] \int {\frac{x^2}{x^2+1}}\,dx = \int {\frac{x^2+1-1}{x^2+1}}\,dx = \int {\frac{x^2+1}{x^2+1}}\,dx - \int {\frac{dx}{x^2+1}}\ [/itex]

- #5

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$$ \int \frac{dx}{1+x^4} $$

$$ x=\sqrt{u}, \quad dx = \frac{1}{2\sqrt{u}} $$

$$ \frac{1}{2} \int \frac{du}{ (1+u^2) \sqrt{u}} $$

$$u = \tan \theta, \quad du = \sec^2 \theta \, d\theta $$

$$ \frac{1}{2} \int \frac{\sec^2\theta}{\sec^2 \theta \sqrt{\tan\theta}} \, d\theta $$

$$ \frac{1}{2} \int \sqrt{ \frac{\cos\theta}{\sin\theta} } \, d\theta $$

Any hints?

- #6

Infrared

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Curious3141

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For the first one, I did some manipulations and made the substitution u=e^e^x, which made it just an indefinite integral of du. Thus, my answer is e^e^x + C

This is fine, but the sub ##u = e^x## works fine and is probably easier to "see".

For the fourth one I did a partial frac decomp and a trig substitution. Final answer:

[tex] x - \arctan(x) + C [/tex]

Good, but you should aim to be able to recognise the easy way (as pointed out by HS-Scientist) immediately in this sort of problem.

For some reason, I don't see anything that I can do with the second and third ones.

Use the sub ##u = e^x## in both. Then partial fractions. For the third, you may need to spot that "easy way" to get the numerator into a lower degree than the denominator in order to use partial fractions.

- #8

Infrared

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This is fine, but the sub ##u = e^x## works fine and is probably easier to "see".

Good, but you should aim to be able to recognise the easy way (as pointed out by HS-Scientist) immediately in this sort of problem.

Use the sub ##u = e^x## in both. Then partial fractions. For the third, you may need to spot that "easy way" to get the numerator into a lower degree than the denominator in order to use partial fractions.

I think that you are missing the quickest way as well. Neither of those problems require partial fractions, though I can certainly see how you used partial fractions for the second.

- #9

pwsnafu

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I think that you are missing the quickest way as well. Neither of those problems require partial fractions, though I can certainly see how you used partial fractions for the second.

Translation: Re-read post #4.

- #10

Curious3141

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I think that you are missing the quickest way as well. Neither of those problems require partial fractions, though I can certainly see how you used partial fractions for the second.

You're right - there's a much quicker way to do both of these.

Took me a while to see it for the second, but then it hit me that after that "trick", the derivative of the denominator equals the numerator.

As for the third - funnily enough, I'd toyed with the idea of using ##(e^x + 1)^2## initially, but abandoned it for no good reason. Makes things so much easier.

OK, if the TS really wants challenging but ultimately elementary integrals, he should work on ##\displaystyle \int \sec\theta d\theta##. If he's not worn out by that, he should try out ##\displaystyle \int \sec^3\theta d\theta##.

Yes, I'm a sadist.

- #11

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Sorry, I already know the trick to integrating the secant, but I'll work on sec^3 theta.

- #12

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$$ \int \sec^3\theta \, d\theta= \frac{1}{2}\ln(\sec\theta + \tan\theta) + \frac{1}{2} \theta\sec\theta + C $$

I'll explain my answer when I get somewhere where I can use my laptop (I did that on my phone)

- #13

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My all-time favorite integral would have to be:

[itex]\displaystyle \int \sqrt{x+k \over x}\,dx[/itex]

It's challenging, frustrating and time-consuming.

[itex]\displaystyle \int \sqrt{x+k \over x}\,dx[/itex]

It's challenging, frustrating and time-consuming.

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- #14

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Hi. Actually, I think that a few of the integrals in this thread are actually pretty simple if we recognize complex numbers.You're right - there's a much quicker way to do both of these.

Took me a while to see it for the second, but then it hit me that after that "trick", the derivative of the denominator equals the numerator.

As for the third - funnily enough, I'd toyed with the idea of using ##(e^x + 1)^2## initially, but abandoned it for no good reason. Makes things so much easier.

OK, if the TS really wants challenging but ultimately elementary integrals, he should work on ##\displaystyle \int \sec\theta d\theta##. If he's not worn out by that, he should try out ##\displaystyle \int \sec^3\theta d\theta##.

Yes, I'm a sadist.

For example, because ##e^{ix} = \cos x + i\sin x## and ##e^{-ix} = \cos x - i\sin x##, we can define sine and cosine (and thus, all trig functions) in terms of complex exponentials.

##\displaystyle \int \sec\theta \ d\theta = \int \frac{2}{e^{i\theta}+e^{-i\theta}} \ d\theta = 2\int \frac{e^{i\theta}}{e^{2i\theta}+1} \ d\theta = -2i\arctan(e^{i\theta}) + C## for some constant C.

- #15

pwsnafu

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As for the third - funnily enough, I'd toyed with the idea of using ##(e^x + 1)^2## initially, but abandoned it for no good reason. Makes things so much easier.

You don't need that either. Just ##u = e^x## and then use the trick on what you get.

Edit:

Re: Mandelbroth

When people ask for these types of question they expect that you reduce the solution so that there are no imaginary expressions, to illustrate that the answer is in fact real.

- #16

Curious3141

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You don't need that either. Just ##u = e^x## and then use the trick on what you get.

I don't know if that's necessarily neater than the method I came up with.

Mine goes like this:

##\displaystyle \int \frac{e^{2x}}{e^x + 1}dx = \int \frac{(e^x + 1)^2 - 2e^x - 1}{e^x + 1} = \int [e^x + 1 - \frac{e^x}{e^x + 1} - 1]dx = \int [e^x - \frac{e^x}{e^x + 1}]dx = e^x - \ln(e^x + 1) + constant##

OK, there's an implicit sub involved in one step, but it could just as easily be seen as recognising an integrand of the form ##\frac{f'(x)}{f(x)}##.

But if we're looking for "clever insights", I still think this way is neater than a sub from the start.

- #17

Curious3141

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My all-time favorite integral would have to be:

[itex]\displaystyle \int \sqrt{x+k \over x}\,dx[/itex]

It's challenging, frustrating and time-consuming.

A simple trig sub reduces this to the integral of the cube of the secant, which is well known (I'd set it as a problem earlier in this thread).

Of course, there could be a more insightful way to do this.

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- #18

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[itex]\displaystyle \int \sqrt{x+k \over x}\,dx[/itex]

It's challenging, frustrating and time-consuming.

I got:

$$2k( \frac{1}{3} (\arctan\arctan\sqrt{x})^3 + \arctan \arctan \sqrt{x}) + C$$

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