# Best method to solve this integral?

1. Mar 26, 2006

### G01

$$\int\frac{dx}{x^2\sqrt{4x+1}}$$ Can someone give me a hint? I've been working on this forever.

2. Mar 26, 2006

### benorin

Let $$u=\sqrt{4x+1} \Rightarrow du=\frac{4dx}{\sqrt{4x+1}}$$
also, $$\frac{1}{x^2}=\frac{16}{(u^2-1)^2}$$
then use trig substitution

3. Mar 26, 2006

### leon1127

NVM, my calculation was wrong. but i can tell you that the answer involves arctanh.

Last edited: Mar 26, 2006
4. Mar 26, 2006

### benorin

This will give

$$\int\frac{dx}{x^2\sqrt{4x+1}} = 4\int \frac{du}{(u^2-1)^2}$$

5. Mar 26, 2006

### G01

ok using trig substitution, i got the integral of cos^2x / sin^3x dx. Is this correct so far?

6. Mar 26, 2006

### G01

OK i think i may have gotten this. Is this the answer.

$$\frac{-\sqrt{4x + 1}}{2x} - 2\ln|\frac{\sqrt{4x + 1} - 1}{\sqrt{4x}}| +C$$

7. Mar 26, 2006

### benorin

I get

$$\int\frac{dx}{x^2\sqrt{4x+1}} = \log \left| \frac{\sqrt{4x+1}+1}{\sqrt{4x+1}-1}\right| -\frac{\sqrt{4x+1}}{2x}+C$$

you can always differentiate to check.

EDIT: I forgot the factor of 4.

Last edited: Mar 26, 2006
8. Mar 26, 2006

### benorin

EDIT: Oops! should be:

Let $$u=\sqrt{4x+1} \Rightarrow du=\frac{1}{2}\frac{4dx}{\sqrt{4x+1}}$$
also, $$\frac{1}{x^2}=\frac{16}{(u^2-1)^2}$$

to give

$$\int\frac{dx}{x^2\sqrt{4x+1}} = 8\int \frac{du}{(u^2-1)^2}= 2\log \left| \frac{\sqrt{4x+1}+1}{\sqrt{4x+1}-1}\right| -\frac{\sqrt{4x+1}}{x}+C$$

Last edited: Mar 26, 2006
9. Mar 26, 2006

### benorin

BTW, for real x,

$$\mbox{arctanh}^{-1}(x) = \frac{1}{2}\log\left| \frac{x+1}{x-1}\right|$$

which would help if you went to www.integrals.com

10. Mar 27, 2006

### G01

$$8\int \frac{du}{(u^2-1)^2}$$

Im screwing up this integral somwhere. Did you use u = sec$$\theta$$? And did you end up with:

$$-4\csc\theta\cot\theta - 4\ln|\csc\theta - \cot\theta| + C$$

Thanks alot for all of this help.

11. Mar 27, 2006

### benorin

You didn't screw-up, I did. My bad, do this rather:

$$\frac{8}{(u^2-1)^2}=\frac{8}{(u-1)^2(u+1)^2}$$

and now partial fractions

$$\frac{8}{(u-1)^2(u+1)^2}=\frac{A}{u-1}+\frac{B}{(u-1)^2}+\frac{C}{u+1}+\frac{D}{(u+1)^2}$$

cross-multiply to get

$$8=A(u-1)(u+1)^2+B(u+1)^2+C(u-1)^2(u+1)+D(u-1)^2$$

plug-in u=1 to get 8=4B or B=2;

plug-in u=-1 to get 8=4D or D=2;

plug-in u=0 to get 8=-A+B+C+D, but B=D=2, so 4=-A+C

plug-in u=2 to get 8=9A+9B+3C+D, but B=D=2, so -12=9A+3C
solving these two equations gives A=-2 and C=2. Finally, we get

$$\int\frac{8du}{(u-1)^2(u+1)^2}=\int\left(\frac{-2}{u-1}+\frac{2}{(u-1)^2}+\frac{2}{u+1}+\frac{2}{(u+1)^2}\right) du = -2\log |u-1|-\frac{2}{u-1}+2\log |u+1|-\frac{2}{u+1}+C$$
$$= 2\log \left| \frac{u+1}{u-1}\right|-\frac{4u}{u^2-1}+C$$

but $$u=\sqrt{4x+1}$$ so...

12. Mar 27, 2006

### G01

ok thanks alot man....