# Integration by Parts

1. Mar 5, 2008

### Goldenwind

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$\int u*dv = uv - \int v*du$$

Last edited: Mar 5, 2008
2. Mar 5, 2008

### Dick

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

3. Mar 5, 2008

### Goldenwind

Did another one here:

Question 4)
$$\int x log_{10}x dx = \int x \frac{ln(x)}{ln(10)} dx$$
$$= \frac{1}{ln(10)}\int xln(x) dx$$
$$= \frac{1}{ln(10)}\int D(\frac{x^2}{2})ln(x) dx$$
$$= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} D(ln(x))dx\right)$$
$$= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} * \frac{1}{x}dx\right)$$
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?

4. Mar 5, 2008

### HallsofIvy

Staff Emeritus
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'".

5. Mar 5, 2008

### Goldenwind

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))??

6. Mar 5, 2008

### Dick

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

7. Mar 5, 2008

### Goldenwind

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

Thanks :)

8. Mar 5, 2008

### 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.

9. Mar 5, 2008

### HallsofIvy

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

10. Mar 5, 2008

### Tedjn

It is from the more general rule that

$$\int \frac{f'(x)}{f(x)}dx = ln|f(x)| + C$$

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

11. Mar 5, 2008

### 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?

12. Mar 5, 2008

### 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.

13. Mar 5, 2008

### 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...