I need some help with integrating {1 dx / (x^2+3)^(3/2) }. I got
(1/3)[ln {(x^2+3)^(1/2) +x} - ln {(3)^(1/2)}] + C. I tried to differentiate this but could not get it back to original form.

2. $\int\frac1{(x^2+3)^{3/2}}~dx$

Did you try the substitution $x=\sqrt3\tan{u}$ ?

3. Originally Posted by Possible actuary
I need some help with integrating {1 dx / (x^2+3)^(3/2) }. I got
(1/3)[ln {(x^2+3)^(1/2) +x} - ln {(3)^(1/2)}] + C. I tried to differentiate this but could not get it back to original form.
$\int \frac{1}{(x^2+3)^{3/2}} \ dx$

I am going to do this in a strange way:
$\int \frac{1}{(x^2+1)^{3/2}} dx = \int \frac{1}{\sqrt{x^2+1}}\cdot \frac{1}{x^2+1} \ dx$

Use substitution, $t = \tan^{-1} x \Rightarrow t' = \frac{1}{x^2+1}$
With $\sec t = \frac{1}{\sqrt{x^2+1}}$

Thus,
$\int \sec t dt = \ln |\sec t + \tan t| + C$

Thus,
$\ln | \sec (\tan^{-1} x) + \tan (\tan^{-1} x) | + C = \ln | \sqrt{x^2+1} + x | + C$

Apparently this can also be done with inverse hyperbolic functions, but I think it is a waste to memorize them. So I did it the standard way.

Now given,
$\int \frac{1}{(x^2+3)^{3/2}} dx = \frac{1}{3^{3/2}} \cdot \int \frac{1}{\left( \frac{x^2}{3} + 1 \right)^{3/2}} \ dx = \frac{1}{3^{3/2}} \cdot \int \frac{1}{\left[ \left( \frac{x}{\sqrt{3}}\right)^2 + 1 \right]^{3/2}} \ dx$

Use the substitution $u= \frac{x}{\sqrt{3}}$ and above work to get:

$\frac{1}{3^{3/2}} \cdot \frac{\sqrt{3}}{1} \cdot \ln \left| \sqrt{ \frac{x^2}{3} + 1} + \frac{x}{\sqrt{3}} \right| + C = \frac{1}{3} \cdot \ln \left| \frac{\sqrt{x^2+3}}{\sqrt{3}} + \frac{x}{\sqrt{3}} \right|+C$

Use the properties of the natural logarithm,
$\frac{1}{3} \cdot \left( \ln | \sqrt{x^2+3} + x | - \ln \frac{1}{\sqrt{3}} \right)+C$

Combine constants,
$\frac{1}{3}\cdot \ln |\sqrt{x^2+3} + x | + C$

4. Yes I did.

[PHP][/PHP]

5. Set $x=\sqrt3\tan{u}\implies{dx=\sqrt3\sec^2u~du}$

The integral becomes to

$\int\frac1{(x^2+3)^{3/2}}~dx=\frac13\int\cos{u}~du=\frac13\sin{u}+k=\fra c{x}{3\sqrt{x^2+3}}+k,~~k\in\mathbb{R}$

6. I think that the use of logarithmic properties confused me. I know that ln (a/b) = ln a - ln b. I am not for sure on ln a - ln (1/b).

That post disappeared. Oh, well.

7. I dunno why The Perfect Hacker deleted his post

Anyway, did you understand my solution?

8. Originally Posted by Krizalid
I dunno why The Perfect Hacker deleted his post

Anyway, did you understand my solution?
I can restore it.

But when I was posting I was doing it without much concentration. So it is possible I made a big mistake. Since your answer does not agree with mine I assumed I made a mistake.

9. Originally Posted by ThePerfectHacker
I can restore it.

But when I was posting I was doing it without much concentration. So it is possible I made a big mistake. Since your answer does not agree with mine I assumed I made a mistake.
I think that we had the same answer up till (1/3)[ ln {(x^2+3)^(1/2) + x / sqrt 3} + C. Then you had put (1/3)[ ln {(x^2+3)^(1/2) +x} - ln (1/sqrt 3) when I thought that the last part should be ln (sqrt 3).

10. Originally Posted by Krizalid
Set $x=\sqrt3\tan{u}\implies{dx=\sqrt3\sec^2u~du}$

The integral becomes to

$\int\frac1{(x^2+3)^{3/2}}~dx=\frac13\int\cos{u}~du=\frac13\sin{u}+k=\fra c{x}{3\sqrt{x^2+3}}+k,~~k\in\mathbb{R}$
I don't know where you pulled k from or how it is figured.

11. Originally Posted by Possible actuary
I don't know where you pulled k from or how it is figured.
it's just the arbitrary constant we get from integrating. we usually call it "C"

12. Originally Posted by Possible actuary
I think that we had the same answer up till (1/3)[ ln {(x^2+3)^(1/2) + x / sqrt 3} + C. Then you had put (1/3)[ ln {(x^2+3)^(1/2) +x} - ln (1/sqrt 3) when I thought that the last part should be ln (sqrt 3).
Okay I restored it. See if it is correct. Because this Whiskey is making it hard to concentrate on the integral. (Do not drink and derive!)

13. Originally Posted by Jhevon
it's just the arbitrary constant we get from integrating. we usually call it "C"
Okay, I know it as C. Thanks!