# Three integrals

1. Jun 11, 2008

### dirk_mec1

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

$$1)\ \int \frac{1}{(x^2+4)^3}\ \mbox{d}x$$

$$2)\ \int \frac{x}{x^4+x^2+1}\ \mbox{d}x$$

$$3)\ \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x$$

3. The attempt at a solution

I got hints but I don't know how to use them.

1) use trigonometric substitution and double angle formulas.
Do I have to use $$u=cos( \theta)$$ and after substituting the double angle formulas what to do next?

2) complete square in numerator
But I'm missing a x^2 term!

3) substitute y-1 and split integrand
Split the term?

2. Jun 11, 2008

### rootX

1. I would say use x = tan(u)

plug in values for x and dx.. and simplify. Integrate with respect to u and then convert from u to x

3. Jun 11, 2008

### dirk_mec1

That gives a $$\frac{1}{ \sec^4(u) }$$ does this makes life easier?

4. Jun 11, 2008

### armis

I think the 2nd one can be solved as follows:

$${\int}xdx/(x^2+1) = 1/2\sdot{\int}dx^2/(x^2+1)$$ and the rest is a piece of cake or am I missing something?

5. Jun 11, 2008

### dirk_mec1

Yes a x^4 -term.

6. Jun 11, 2008

### tiny-tim

Hi dirk_mec1!

1) Hint: $$\frac{1}{ \sec^4(u) }\,=\,cos^4u$$

2) Hint: No … you're missing a units term.

3) Do the substitution anyway, and show us what you get.

7. Jun 11, 2008

### armis

The last one:

$$3)\ \int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = {\int}xde^x/(x+1)^2 = {\int} (x+1)de^x/(x+1)^2 - de^x/(x+1)^2$$

The rest I leave for you. If I am not mistaken you should get $$e^x/(x+1)$$

Sorry for my LaTeX skills. I haven't figured it out yet

8. Jun 11, 2008

### dirk_mec1

Hi tiny tim,

1) Oh now I see how to do the first one!
2) what is a units term?
3) working on it right now.

9. Jun 11, 2008

### tiny-tim

Sorry … I meant an ordinary number (with no x).

Subtract a number from the denominator, to make a perfect square.

(armis, nice LaTeX, but just give hints … unless the OP didn't follow the last hint! )

10. Jun 11, 2008

### armis

yeah, sorry tiny-tim. Although often times it's kind of tricky to express my thoughts in english just verbaly but I'll try harder next time. Thanks god math notation is the same for everyone

By the way I know it's not the right place to ask but how does one use the hide feature?

11. Jun 11, 2008

### dirk_mec1

You mean like: $$(x^2+1)^2 -x^2$$

Use \frac{}{} for fractions

12. Jun 11, 2008

### armis

oh, I see. Thanks

13. Jun 11, 2008

### tiny-tim

No … like $$(x^2+1)^2 - 3$$ (but different, of course!)
hmm … I was going to ask "what hide feature?" … and then I vaguely remembered seeing something about it … I think you type the full answer, but with a tag which hides it unless someone puts the mouse over it.

But I've no idea how to operate it.

You'd better ask someone more knowledgable than me!

Anyway, it's good practice only giving hints!

14. Jun 12, 2008

### dirk_mec1

So you mean:

$$\left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}$$

So we get

$$\int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x$$

Now...what to do next?

Last edited: Jun 12, 2008
15. Jun 12, 2008

### tiny-tim

Woohoo!
Make the obvious substitution.

16. Jun 12, 2008

### dirk_mec1

Let $$u = x^2+\frac{1}{2}$$

Then

$$\int \frac{x}{x^4+x^2+1}\ \mbox{d}x = \int \frac{x}{ \left( x^2 + \frac{1}{2} \right)^2 +\frac{3}{4}}\ \mbox{d}x = \frac{1}{2} \cdot \int \frac{1}{u^2+\frac{3}{4}}\ \mbox{d}x = \frac{1}{2} \arctan \left( \frac{u}{\frac{1}{2} \sqrt{3}} \right)\ +\ C$$

Now put the u in terms of x back and we've got the answer. Is this okay Tim?

17. Jun 12, 2008

### tiny-tim

Yes, that's fine … except you've left out the (2/√3) outside the arctan.

(See Hootenanny's Library entry, List of Standard Integrals)

How did you get on with question 3?

18. Jun 12, 2008

### armis

I am sorry to interrupt, but wasn't there a much easier way to solve the 2nd problem?

$${\int}\frac{xdx}{(x^2+1)} = 1/2\sdot{\int}\frac{dx^2}{(x^2+1)}$$

19. Jun 12, 2008

### dirk_mec1

Whoops, you're right.

I use the wiki-page

$$\int \frac{x e^x}{(x+1)^2}\ \mbox{d}x = \frac{x e^x}{(x+1)^2} - \int e^x \frac{(x+1)^2-2x^2-2x}{(x+1)^4}$$

This is getting nasty....

By the way I didn't understand Armis' hint

Last edited: Jun 12, 2008
20. Jun 12, 2008

### dirk_mec1

Where did you get this integral? This isn't the second integral I posted, right?