Integration

[pinkcrushin22]

How do I integrate 3/(1+2sqrt(x)). (Using u-substitution) I have tried several things to solve this problem and I am so frustrated with it. It seemed like an easy problem and obvious. Your u=1+2sqrt(x), but when you take the dervitive of this you get du=x^-1/2= 1/sqrt(x)dx. This gets me know where. I found the answer I was wondering if someone could please show me how you get to this answer 3/2(2sqrt(x)-log(2sqrt(x+1) + 1). Thanks I would really appreciate it.

 

[daon2]

How do I integrate 3/(1+2sqrt(x)). (Using u-substitution) I have tried several things to solve this problem and I am so frustrated with it. It seemed like an easy problem and obvious. Your u=1+2sqrt(x), but when you take the dervitive of this you get du=x^-1/2= 1/sqrt(x)dx. This gets me know where. I found the answer I was wondering if someone could please show me how you get to this answer 3/2(2sqrt(x)-log(2sqrt(x+1) + 1). Thanks I would really appreciate it.

You were on a good path. Solve for \(\displaystyle dx=\sqrt{x}du\). So you now have as your integrand \(\displaystyle \dfrac{1}{u}dx = \dfrac{1}{u}\sqrt{x} du\).

Now you can get \(\displaystyle \sqrt{x}\) in terms of \(\displaystyle u\) using your original substitution, and replacing it in your integrand should make it pretty clear how to proceed. \(\displaystyle \sqrt{x}=\dfrac{u-1}{2}\).

 

[pinkcrushin22]

You were on a good path. Solve for \(\displaystyle dx=\sqrt{x}du\). So you now have as your integrand \(\displaystyle \dfrac{1}{u}dx = \dfrac{1}{u}\sqrt{x} du\).

Now you can get \(\displaystyle \sqrt{x}\) in terms of \(\displaystyle u\) using your original substitution, and replacing it in your integrand should make it pretty clear how to proceed. \(\displaystyle \sqrt{x}=\dfrac{u-1}{2}\).

Thank you this was really helpful!!!!! I had never had anyone show me that before. I didn't even know you can do that. I really appreciate this. Thank you for your help!!!!!

 

[pinkcrushin22]

one more question

Thanks again for the help. Now after the sqrt(x) = u-1/2 we take that and plug it in.
This gives me the integral of (1/u) (u-1/2)
Then put them together and get = u-1/u
which then you get after you integrate this 3/2(1/2u^2-u*u)

I went somewhere wrong or I misunderstood.

 

[JeffM]

Thanks again for the help. Now after the sqrt(x) = u-1/2 we take that and plug it in.
This gives me the integral of (1/u) (u-1/2)
Then put them together and get = u-1/u
which then you get after you integrate this 3/2(1/2u^2-u*u)

I went somewhere wrong or I misunderstood.

\(\displaystyle \int \dfrac{3}{1 + 2\sqrt{x}} dx.\) This is the original problem.

\(\displaystyle u = 1 + 2\sqrt{x} = 1 + 2x^{(1/2)} \implies \dfrac{du}{dx} = 0 + \frac{1}{2} * 2x^{\{(1/2) - 1\}} = x^{-(1/2)} = \dfrac{1}{\sqrt{x}} \implies dx = \sqrt{x} * du.\) Do you see how daon got here?

\(\displaystyle u = 1 + 2\sqrt{x} \implies u - 1 = 2\sqrt{x} \implies \sqrt{x} = \dfrac{u - 1}{2} \implies dx = \dfrac{u - 1}{2} * du.\)

Do you see how we get here?

You can now replace the function in x with the corresponding function in u AND replace dx with its equivalent in terms of du. This is the whole point of u-substitution, replace the function in x and dx with a function in u and du.

\(\displaystyle \int \dfrac{3}{1 + 2\sqrt{x}} dx = \int \dfrac{3}{u} * \dfrac{u - 1}{2} du = \dfrac{3}{2}\int \left(1 - \dfrac{1}{u}\right)du.\)

Proceed

 

[pinkcrushin22]

\(\displaystyle \int \dfrac{3}{1 + 2\sqrt{x}} dx.\) This is the original problem.

\(\displaystyle u = 1 + 2\sqrt{x} = 1 + 2x^{(1/2)} \implies \dfrac{du}{dx} = 0 + \frac{1}{2} * 2x^{\{(1/2) - 1\}} = x^{-(1/2)} = \dfrac{1}{\sqrt{x}} \implies dx = \sqrt{x} * du.\) Do you see how daon got here?

\(\displaystyle u = 1 + 2\sqrt{x} \implies u - 1 = 2\sqrt{x} \implies \sqrt{x} = \dfrac{u - 1}{2} \implies dx = \dfrac{u - 1}{2} * du.\)

Do you see how we get here?

You can now replace the function in x with the corresponding function in u AND replace dx with its equivalent in terms of du. This is the whole point of u-substitution, replace the function in x and dx with a function in u and du.

\(\displaystyle \int \dfrac{3}{1 + 2\sqrt{x}} dx = \int \dfrac{3}{u} * \dfrac{u - 1}{2} du = \dfrac{3}{2}\int \left(1 - \dfrac{1}{u}\right)du.\)

Proceed

I got how daon got there its was after that. Thank you now I get it!!!!!