Please help integrate...

**Possible actuary** · June 16th 2007, 06:57 PM

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.

**Krizalid** · June 16th 2007, 07:04 PM

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

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

**ThePerfectHacker** · June 16th 2007, 07:12 PM

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:
Instead find,
$\int \frac{1}{(x^2+1)^{3/2}} dx = \int \frac{1}{\sqrt{x^2+1}}\cdot \frac{1}{x^2+1} \ dx$

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

**Possible actuary** · June 16th 2007, 07:23 PM

Yes I did.

[PHP][/PHP]

**Krizalid** · June 16th 2007, 07:23 PM

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

The integral becomes to

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

**Possible actuary** · June 16th 2007, 07:26 PM

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.

**Krizalid** · June 16th 2007, 07:48 PM

I dunno why The Perfect Hacker deleted his post

Anyway, did you understand my solution?

**ThePerfectHacker** · June 16th 2007, 07:50 PM

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.

**Possible actuary** · June 16th 2007, 08:07 PM

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).

**Possible actuary** · June 16th 2007, 08:08 PM

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.

**Jhevon** · June 16th 2007, 08:10 PM

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"

**ThePerfectHacker** · June 16th 2007, 08:10 PM

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!)

**Possible actuary** · June 17th 2007, 06:18 AM

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!

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