# Problem with integral

1. Oct 8, 2005

### sony

Ok, so I have the integral of (x^2 + 1) / (x^3 + 8) dx

I use partial fractions and end up with two integrals, the one I cannot solve is:

1/12 * "integral of" (7x-4) / [(x-1)^2 +3 ] dx
With u=x-1 I get
(7u-3) / (u^2+3) du

But I have no Idea of how to solve it, the answer in the books shows the step up to this, but jumps over what to do with the last part and writes the answer (something involving tan^-1 and ln)

Thanks

2. Oct 8, 2005

### amcavoy

Factor out the 3 in the denominator:

$$\int\frac{7u-3}{u^2+3}\,du=\frac{1}{3}\int\frac{7u-3}{\left(\frac{u}{\sqrt{3}}\right)^2+1}\,du$$

Then make the subsititution $u=\sqrt{3}\tan{\theta}$.

Alex

3. Oct 9, 2005

Oh, thanks!

4. Oct 9, 2005

### sony

Ok I'm very confused about the doing substitution _two_ times, can someone please go threw every step of this?

5. Oct 9, 2005

### arildno

Fra børjan av???

6. Oct 9, 2005

### sony

Sorry, nevermind! I think I can figure this out, can someone please just tell me what the integral of Tan[x] equals?

7. Oct 9, 2005

### arildno

It is C-ln(|cos(x)|), where C is an integration constant.

8. Oct 9, 2005

### sony

Ok, I see it now. Took some time :P

Thanks!

9. Oct 9, 2005

### sony

Ok, bah. I don't get it. Can someone please take it from what apmcavoy wrote (start with subsitution of u)

10. Oct 9, 2005

### arildno

You have then:
$$du=\frac{\sqrt{3}d\theta}{\cos^{2}\theta}$$
Since $$tan^{2}\theta+1=\frac{1}{\cos^{2}\theta}$$
we get:
$$\frac{1}{3}\int\frac{7u-3}{(\frac{u}{\sqrt{3}})^{2}+1}du=\frac{1}{3}\int\sqrt{3}(7tan\theta-3)d\theta$$
All right?

11. Oct 9, 2005

### sony

Yeah, thanks. I made it to the last step, but I dont see what happens with the Cos bit... :P

(EDIT: The cos bit in the solution of the integral of Tan phi)

12. Oct 9, 2005

### arildno

Okay, you'll basically need to rearrange an expression like cos(Atan(y)), which will appear within the logarithm.

In order to do this, not that by definition of the Atan and tan functions, we have:
$$y=\tan{Atan(y)}=\frac{\sin(Atan(y))}{\cos(Atan(y))}=\frac{\sqrt{1-\cos^{2}(Atan(y))}}{\cos(Atan(y))}(1)$$
where I've used the identity $\sin^{2}x=1-\cos^{2}x$ for all x.

Thus, from (1), we get $$y^{2}\cos^{2}(Atan(y))=1-\cos^{2}(Atan(y))$$
by which we have:
$$|\cos(Atan(y))|=\frac{1}{\sqrt{1+y^{2}}}$$

Thus, we have rewritten the troublesome expression, in that we now have:
$$ln|\cos(Atan(y))|=-\frac{1}{2}ln(y^{2}+1)$$

13. Oct 9, 2005

### sony

Ok, this made it clear. Thank you!

14. Oct 9, 2005

### arildno

You're welcome.