1. ## Definite integral

How do I find the definite integral of 1 / (x^1/2 + 2x + x^3/2) dx, with upper limit of 4 and lower limit of 0? The hint says that I need to factor out the denominator, with which I got x^1/2 times (1 + 2x^1/2 + x). What do I do from here, and would u/du substitution be the best way? Thanks!

2. ## Re: Definite integral

Originally Posted by Hamzawesome
How do I find the definite integral of 1 / (x^1/2 + 2x + x^3/2) dx, with upper limit of 4 and lower limit of 0? The hint says that I need to factor out the denominator, with which I got x^1/2 times (1 + 2x^1/2 + x). What do I do from here, and would u/du substitution be the best way? Thanks!
Put $\displaystyle u=\sqrt{x}$ then simplify and use partial fractions.

CB

3. ## Re: Definite integral

Originally Posted by Hamzawesome
How do I find the definite integral of 1 / (x^1/2 + 2x + x^3/2) dx, with upper limit of 4 and lower limit of 0? The hint says that I need to factor out the denominator, with which I got x^1/2 times (1 + 2x^1/2 + x). What do I do from here, and would u/du substitution be the best way? Thanks!
$\displaystyle \sqrt{x}=t \Rightarrow \frac{1}{2\sqrt{x}}dx=dt$

$\displaystyle I=2\cdot\int\limits_{0}^{2} \frac{1}{t^2+2t+1} \, dt =2\cdot\int\limits_{0}^{2} \frac{1}{(t+1)^2} \, dt$

4. ## Re: Definite integral

Originally Posted by princeps
$\displaystyle \sqrt{x}=t \Rightarrow \frac{1}{2\sqrt{x}}dx=dt$

$\displaystyle I=2\cdot\int\limits_{0}^{2} \frac{1}{t^2+2t+1} \, dt =2\cdot\int\limits_{0}^{2} \frac{1}{(t+1)^2} \, dt$
Why did you put for the interval of the integral (2,0)? In the problem the upper limit is 4 and the lower limit is 0, did you put 2 and 0 for the limits by mistake or was that supposed to be there?

5. ## Re: Definite integral

Originally Posted by Hamzawesome
Why did you put for the interval of the integral (2,0)? In the problem the upper limit is 4 and the lower limit is 0, did you put 2 and 0 for the limits by mistake or was that supposed to be there?
Since substitution is $\displaystyle \sqrt {x}=t$ you have to change limits .

$\displaystyle \sqrt{0}=0$ and $\displaystyle \sqrt{4}=2$

6. ## Re: Definite integral

Originally Posted by princeps
Since substitution is $\displaystyle \sqrt {x}=t$ you have to change limits .

$\displaystyle \sqrt{0}=0$ and $\displaystyle \sqrt{4}=2$
After that I used u/du substitution for that last integral you included; u = t + 1 and du = dt, so the integral was 2int(2,0) du/u^2 which ended up as [-2/u](2,0) and then [-2/t+1](2,0) so for the final answers I got 4/3.

7. ## Re: Definite integral

Originally Posted by Hamzawesome
After that I used u/du substitution for that last integral you included; u = t + 1 and du = dt, so the integral was 2int(2,0) du/u^2 which ended up as [-2/u](2,0) and then [-2/t+1](2,0) so for the final answers I got 4/3.
That should be a correct answer...

8. ## Re: Definite integral

Originally Posted by princeps
That should be a correct answer...
I hope you don't mean there are more than one....lol. Thanks, regardless.

9. ## Re: Definite integral

interesting to note that the original definite integral is improper ...