sustitution intregration again

**startingover** · June 7th 2007, 04:27 AM

I think I am getting closer, but necessarily correct.

integral (10x^5-5x)/square root (x^4-x^2+6) dx

let x = x^4 -x^2 +6
d/dx = 4x^3 - 2x or 2x(2x^2 -1)

1/2 integral x^-1/2 du
1/2 *(u^1/2)/1/2 + C
1/4*u^1/2 + C
1/4 (x^4-x^2+6)^1/2 +C

**ThePerfectHacker** · June 7th 2007, 06:42 AM

Originally Posted by startingover

I think I am getting closer, but necessarily correct.

integral (10x^5-5x)/square root (x^4-x^2+6) dx

Check you integral again. I think you made a mistake on it.

**startingover** · June 7th 2007, 06:57 AM

THP:
Man, you are good....you are right, I didn't write down the problem correctly.

integral (10x^3-5x)/square root (x^4-x^2+6) dx

**Pterid** · June 7th 2007, 07:16 AM

To find $I = \int \frac{10x^3-5x}{\sqrt{x^4-x^2+6}} dx = \frac{5}{2}\int \frac{4x^3-2x}{\sqrt{x^4-x^2+6}} dx$ :

Substitute $u = x^4 - x^2 + 6~~~\Rightarrow~~~du = (4x^3 - 2x) dx$ .

Then $I = \frac{5}{2} \int \frac{1}{\sqrt{u}} du = \frac{5}{4}\sqrt{u} + C = \frac{5}{4}\sqrt{x^4 - x^2 + 6} + C$ .

**startingover** · June 7th 2007, 07:30 AM

The most confusion I face is in figuring out how you get the 5/2...
I understand that (x^4-X^2+6)^-1/2
Then it becomes 1/n+1^n+1
I assume it is because the numerator becomes 5x(2x^2-1)...
Please advise.

**Pterid** · June 7th 2007, 08:34 AM

Originally Posted by startingover

The most confusion I face is in figuring out how you get the 5/2...
I understand that (x^4-X^2+6)^-1/2
Then it becomes 1/n+1^n+1
I assume it is because the numerator becomes 5x(2x^2-1)...
Please advise.

OK. Right at the beginning. I decided to "bring outside" a factor of $5/2$ on the first line. Like this:

$\int \frac{10x^3-5x}{\sqrt{x^4-x^2+6}} dx ~~=~~ \frac{5}{2} \cdot \int \frac{(\frac{2}{5})\left(10x^3-5x\right)}{\sqrt{x^4-x^2+6}} dx ~~=~~ \frac{5}{2} \cdot \int \frac{(4x^3-2x)}{\sqrt{x^4-x^2+6}} dx$

Hopefully, from that line above, you can see how I did it. The more important question, I guess, is why I did it!

The reason is, because then it's easy to see the substitution you should use.

After you "bring out" the $5/2$ , the remaining integral has a particular, special form:

$\int \frac{(4x^3-2x)}{\sqrt{x^4-x^2+6}} dx ~~=~~ \int f(u) \left( \frac{du}{dx} \right) dx$

Here, $u = x^4 - x^2 + 6;$

$f(u) = \frac{1}{\sqrt{u}} ~=~ \frac{1}{\sqrt{x^4 - x^2 + 6}};$

$\left( \frac{du}{dx}\right) = (4x^3 - 2x)$ .

This whole process of switching (or substituting) $u$ for $x$ is called 'integration by substitution'.

Then, it becomes easy to finish the integral, because you can say:

$\int f(u) \left( \frac{du}{dx} \right) dx ~~=~~ \int f(u) du$

and because you know $f(u) = \frac{1}{\sqrt{u}} = u^{-1/2}$ , you can integrate it using the normal rule:

$\int u^n du ~~=~~ \frac{u^{n+1}}{n+1} + C$ .

- - -

I hope that helps to explain what I was doing. Still, if there's a specific bit you don't understand, I'll "zoom in" on it and explain further.

**Krizalid** · June 7th 2007, 09:59 AM

$\int {\frac{{10x^3 - 5x}}<br /> {{\sqrt {x^4 - x^2 + 6} }}~dx} = 5\int {\frac{{2x^3 - x}}<br /> {{\sqrt {x^4 - x^2 + 6} }}~dx} = \frac{5}<br /> {2}\int {\frac{{4x^3 - 2x}}<br /> {{\sqrt {x^4 - x^2 + 6} }}~dx}$

It's faster if you only take $u=\sqrt{x^4 - x^2 + 6}$

**^_^Engineer_Adam^_^** · June 7th 2007, 07:19 PM

$I = \frac{5}{2} \int \frac{1}{\sqrt{u}}du$
$I = \frac{5}{2} . \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$

$I = \frac{5}{2} . \frac{\sqrt{u}}{\frac{1}{2}}$

$I = 5 . \sqrt{u}$

$I = 5 . \sqrt{x^4-x^2+6}$

**Pterid** · June 8th 2007, 03:01 AM

Originally Posted by ^_^Engineer_Adam^_^

$I = 5 . \sqrt{x^4-x^2+6}$

You are right, I was wrong by that factor of 4.

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