# reconciling intergration by parts with direct substitution

• Feb 6th 2010, 07:12 PM
vince
reconciling intergration by parts with direct substitution
obviously one can use integration by parts to attempt to solve any integral not just those with funky products in the intergrand so with that in mind i was trying to solve the integral,

$\displaystyle \int\frac{1}{xln(x)}dx$

using integration by parts after i had already noticed that direct substituion of u=ln(x) would do the trick.

But with the following choices this is what I'm getting:
Let f(x) = $\displaystyle \frac{dx}{x}$, g(x) = $\displaystyle \frac{1}{ln(x)}$

Then
$\displaystyle \int\frac{1}{xln(x)}dx$ = $\displaystyle \frac{ln(x)}{ln(x)} - \int\frac{ln(x)}{-xln(x)^2}dx$
= $\displaystyle 1 +\int\frac{1}{xln(x)}dx$

As you can see, this is nonsensical.

Would someone please tell me where I'm erring.

Thanks

The correct result is $\displaystyle \int\frac{1}{xln(x)}dx = ln(ln(x)) + C$
• Feb 6th 2010, 07:37 PM
mr fantastic
Quote:

Originally Posted by vince
obviously one can use integration by parts to attempt to solve any integral not just those with funky products in the intergrand so with that in mind i was trying to solve the integral,

$\displaystyle \int\frac{1}{xln(x)}dx$

using integration by parts after i had already noticed that direct substituion of u=ln(x) would do the trick.

But with the following choices this is what I'm getting:
Let f(x) = $\displaystyle \frac{dx}{x}$, g(x) = $\displaystyle \frac{1}{ln(x)}$

Then
$\displaystyle \int\frac{1}{xln(x)}dx$ = $\displaystyle \frac{ln(x)}{ln(x)} - \int\frac{ln(x)}{-xln(x)^2}dx$
= $\displaystyle 1 +\int\frac{1}{xln(x)}dx$

As you can see, this is nonsensical. Mr F says: No it's not. All it says is that the difference between the two integrals is a constant.

Would someone please tell me where I'm erring.

Thanks

The correct result is $\displaystyle \int\frac{1}{xln(x)}dx = ln(ln(x)) + C$

..
• Feb 6th 2010, 07:48 PM
vince
Thanks for the response Mr.F. But rearranging the equation:

http://www.mathhelpforum.com/math-he...20c78976-1.gif = http://www.mathhelpforum.com/math-he...a9658da4-1.gif
= http://www.mathhelpforum.com/math-he...1a804a68-1.gif

yields 0=1!!!

Also, direct substitution shows:
The correct result is http://www.mathhelpforum.com/math-he...25100b9f-1.gif
Clearly non constant function of x.
• Feb 7th 2010, 01:08 AM
HallsofIvy
And that's the whole point! Your calculation does not say that the integral is a constant, it says the the difference between the two integrals is a constant. Integrals of the same function can differ by a constant.

What you wrote"
$\displaystyle \int \frac{1}{x ln(x)} dx= 1+ \int \frac{1}{x ln(x)}dx$

is wrong because you have not included the constant of integration you have in your last formula. What you should have is

$\displaystyle \int \frac{1}{x ln(x)} dx+ C_1= 1+ \int \frac{1}{x ln(x)}dx+ C_2$
where $\displaystyle C_1$ and $\displaystyle C_2$ are the constants of integration. That reduces to the statement that

$\displaystyle C_1= 1+ C_2$
and there is nothing peculiar about that. The two methods just use different constants of integration.
• Feb 7th 2010, 05:41 AM
vince
well this was instructive. all that comes to mind is 'oops' (Wink).
But it's rather sneaky in that those constants would only appear after the evaluation of $\displaystyle \int \frac{1}{x ln(x)}$, which in your explanation is implicit. that is you're anticipating their arrival because you know one can't subtract indefinite intregals from each other without first accounting for the constants. Thanks!

This example gives me a new appreciation for the constants. i ve always framed them as numbers that'd ensure initial conditions would be met, so as to pin down the exact antiderivative from the continuum of antiderivatives to choose from when evaluating indefinite integrals. In this case, they do more than just that: they show that two methods are compatible but that one of them is actually the only way one will get a useful answer. this last point is nothing new, especially when using integration by parts.
thanks again.