1. ## Problem with Integration

Hi

I am having trouble trying to integrate this equation:

$\displaystyle \int \sqrt(1+2x+2x^3)$

how would i approach an expression like this?

Do i integrate the inside and then the outside?

P.S

2. Are you sure it's $\displaystyle \int{\sqrt{1+2x+3x^3}}\;{dx}$?

3. Well, the original question asked to find the length of arc of $\displaystyle f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})$
$\displaystyle y=2-\frac{1}{2}x$ between x=0, y=3

I used the length of arc rule:

$\displaystyle \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

This is where i was stuck:

$\displaystyle \int_0^3 \sqrt{1+2x+2x^3}dx$

4. Originally Posted by Paymemoney
Well, the original question asked to find the length of arc of $\displaystyle f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})$
$\displaystyle y=2-\frac{1}{2}x$ between x=0, y=3

I used the length of arc rule:

$\displaystyle \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

This is where i was stuck:

$\displaystyle \int_0^3 \sqrt{1+2x+2x^3}dx$
Dear Paymemoney,

I have some problems regarding your question. You want to find the arc length of $\displaystyle f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$ is'nt?

What is the second equation $\displaystyle y=2-\frac{1}{2}x$ ??? How did you get this? Do you want to find the arc length of this curve too??

5. Originally Posted by Sudharaka
Dear Paymemoney,

I have some problems regarding your question. You want to find the arc length of $\displaystyle f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$ is'nt?

What is the second equation $\displaystyle y=2-\frac{1}{2}x$ ??? How did you get this? Do you want to find the arc length of this curve too??

opps srry my mistake the equation is only $\displaystyle f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$

6. Originally Posted by Paymemoney
opps srry my mistake the equation is only $\displaystyle f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$
Dear Paymemoney,

You have simplified incorrectly,

$\displaystyle \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

$\displaystyle =\int_0^3\sqrt{1+4x^2(1+x^2)}dx$

$\displaystyle =\int_0^3\sqrt{1+4x^2+4x^4}dx$

$\displaystyle =\int_0^3\sqrt{(1+2x^2)^2}dx$

Can you continue from here??

7. Originally Posted by Sudharaka
Dear Paymemoney,

You have simplified incorrectly,

$\displaystyle \int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

$\displaystyle =\int_0^3\sqrt{1+4x^2(1+x^2)}dx$

$\displaystyle =\int_0^3\sqrt{1+4x^2+4x^4}dx$

$\displaystyle =\int_0^3\sqrt{(1+2x^2)^2}dx$

Can you continue from here??
yes,
thank you, i worked it out