Integration by Parts Homework Help - Q2

[Goldenwind]

Homework Statement

In preparation for an exam next week, I'm solving some of the end-of-chapter questions. There are 30 questions. I've solved a few on my own, but here's one I'm getting stuck on.

Question 2)
\int x5^x dx, problem here is I don't know how to deal with 5^x.
The correct answer should be: \frac{x}{ln5}5^x - \frac{1}{(ln5)^2}5^x + C

To avoid spamming the forum with many threads, I'll just post in this one whenever I hit an integral I'm unsure of. For now, it's just the one above.

Homework Equations

\int u*dv = uv - \int v*du

 

[Dick]

It might look a little easier if you write 5^x=exp(ln(5)*x). Put the exp into the dv part.

 

[Goldenwind]

Did another one here:

Question 4)
\int x log_{10}x dx<br /> = \int x \frac{ln(x)}{ln(10)} dx
= \frac{1}{ln(10)}\int xln(x) dx
= \frac{1}{ln(10)}\int D(\frac{x^2}{2})ln(x) dx
= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} D(ln(x))dx\right)
= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} * \frac{1}{x}dx\right)
And yeah, getting too lazy to TeX it all. Question is, when I finally get rid of all integrals, the +C will be inside the 1/ln(10) bracket. Do I write C/ln(10), or do I assume that whatever C is will already have that applied, so I just write C?

 

[HallsofIvy]

Yes, the constant C/ln(10) is the same as the constant C' - and it isn't even necessary to distinguish between "C" and "C'".

 

[Goldenwind]

It might look a little easier if you write 5^x=exp(ln(5)*x). Put the exp into the dv part.

So, 5^x = exp(ln(5^x)) = exp(xln(5)) = e^(xln(5))... I see that much.
I also know the derivative/antiderivative of e^x = e^x.
What is the derivative/antiderivative of e^(xln(5))??

 

[Dick]

Use the chain rule. The derivative of e^(a*x)=a*e^(a*x). What's the antiderivative?

 

[Goldenwind]

e^(ax) / a, methinks... 'cause then taking the derivative of that would pull down another 'a', canceling out the division. Makes sense.

Thanks :)

 

[Goldenwind]

Why is \int \frac{D(1+x^2)}{1+x^2}dx = ln(1+x^2)?

I know D(ln(x)) = 1/x, however I don't see how my book did this calculation.

 

[HallsofIvy]

The point is that \int\frac{du}{u}= ln|u|+ C, No matrer what "u" is. In this case u= 1+ x².

 

[Tedjn]

It is from the more general rule that

\int \frac{f&#039;(x)}{f(x)}dx = ln|f(x)| + C

which comes from the chain rule. If you think about substituting u for 1 + x², then what is du in terms of dx?

 

[Goldenwind]

If u = 1 + x^2, then du = 2x.
Thanks for the ln|blah| tip! I remember being taught it, but had forgotten.

One final question for tonight.
I'm in the middle of integrating arcsin(x), and as a segment, I need to compute the antiderivative:
\int \frac{x}{\sqrt{1 - x^2}}

From looking up the answer, I see that this integral is equivalent to \sqrt{1 - x^2}. Why would this be? I see how it can be broken up into \frac{x}{1} * (1 - x^2)^{-\frac{1}{2}}. Does this help me, or am I missing another rule?

 

[Tedjn]

Yes, you are right, although the division by 1 is not really necessary. Think chain rule again, and see if you can separate one part of the integral as a derivative of another part, then substitute.

 

[Goldenwind]

Derivative of the top = 1

Derivative of the bottom = -2x * (1 - x^2)^{-\frac{3}{2}}*-\frac{1}{2}
= x(1 - x^2)^{-\frac{3}{2}}

Sorry, but I still don't see it >.<;
I see how the chain rule was used to find the derivative of the bottom, but otherwise...