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By "obvious" I mean something like: $$\int {x^2\over x^3+2}dx$$ I know I can use u-sub with $u=x^3+2$ and $du=3x^2dx$ to get: $${\ln(x^3+2)\over3}$$ But what about a problem like: $$\int 35\sqrt x e^\sqrt x dx$$ I first tried integrating by parts because I thought of it as functions $35\sqrt x$ and $e^\sqrt x$ multiplied together. I didn't know to set $u=\sqrt x$ until I got ${70\over 3}x^{3\over2}-35\int xe^\sqrt x$ through integration by parts (and then looked it up on an online calculator), and I'm assuming that would go into another loop of integration by parts because of the way derivatives with $e$ works.

How do I know whether to try a u-sub or integration by parts first when there isn't an apparent function I can differentiate and easily replace with in the integral? This includes problems with and without $e$.

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    $\begingroup$ On general principles of "substitute so that common terms are unified", I'd have tried the $u = \sqrt{x}$ substitution on that integral. $\endgroup$ – Patrick Stevens May 26 '18 at 7:47
  • $\begingroup$ Thanks for the reply. Can you clarify which common terms you are talking about? $\endgroup$ – VolTorian May 27 '18 at 6:07
  • $\begingroup$ In this instance, $x$ appears only in the form $\sqrt{x}$, so I'd try substituting it away. $\endgroup$ – Patrick Stevens May 27 '18 at 8:27
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In $\int \frac{x^2}{x^3+2}$ you need to take the $u$ substitution of a term that cancels the numerator. Here if we take $u=x^3+2$ then $dx=\frac{du}{3x^2}$. So the $x^2$ term in the numerator gets canceled.

In the cases like $\int(6x^2)(2x^3+5)^6dx$ it is always advisable to take the number with the highest power. In this case $u=2x^3+5$

In the cases like $\int cos(5x-7)dx$ it is always advisable to take the number in the brackets. In this case $u=5x-7$

In the cases like $\int 35\sqrt{x}e^{\sqrt{x}}$ it is always advisable to take the exponent of $e$ as $u$. In this case $u=\sqrt{x}$

Another example of this is $\int \frac{e^x}{e^x}dx=\int1+e^{-x}dx$, even in this case we take the exponent of e as $u$.

$u=-x$

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  • $\begingroup$ So in generally unless it's like my mentioned obvious case, I'd want to set the exponent of $e$ as u? Can you think of other integrals that wouldn't work this way? $\endgroup$ – VolTorian May 29 '18 at 3:06
  • $\begingroup$ Example-$\frac{e^x}{1-e^x}dx$ here we take $u=1-e^x$ $\endgroup$ – tien lee May 29 '18 at 3:11
  • $\begingroup$ Well I would say that's an obvious one with a clear u and du but thanks for the reply anyway $\endgroup$ – VolTorian May 29 '18 at 17:36

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