# 1/(1+sqrt(x)) dx

1. Jan 14, 2010

### Alexx1

How do you solve these 3 integrals? :

Integral 1 : 1/(1+sqrt(x)) dx

Integral 2: (x^3)*(e^x^2)

Integral 3: (x*e^x)/((x+1)^2)

I have no idea how to solve these integrals..

2. Jan 14, 2010

### tiny-tim

Hi Alexx1!

(have an integral: ∫ and a square-root: √ and try using the X2 tag just above the Reply box )
For 1 and 2, use the obvious substitutions.

3. Jan 14, 2010

### Mentallic

Re: Integrals

1) let $1+\sqrt{x}=u$

2) let $x^2=u$ and then apply the integration by parts.

3) Apply integration by parts by letting $u=xe^x$ and $v'=(x+1)^{-2}$

If you're still stuck after this, show us what you've done and we'll help you further. Good luck!

4. Jan 14, 2010

### Alexx1

Re: Integrals

Thx! I found the third one.
But the first and the second one, is it like

u = 1+sqrt(x) --> du = 1/2sqrt(x) dx --> dx = 2sqrt(x) du
u = x^2 --> du = 2xdx --> dx = du/2

?

(I learned to use t = ... --> dt = ... dx)

Last edited by a moderator: Jan 14, 2010
5. Jan 14, 2010

### HallsofIvy

Re: Integrals

and sqrt(x)= u- 1 so dx= 2(u- 1)du

Yes, that's right.

6. Jan 14, 2010

### Alexx1

Re: Integrals

For the first one I become:

integral: 2(u-1)/u du

= 2 (integral u/u du - integral 1/u du)

= 2 (u - ln u)

= 2 (sqrt(x)+1 - ln (sqrt(x)+1))

But the correct answer is: 2 (sqrt(x) - ln (sqrt(x)+1))

What have I done wrong?

7. Jan 14, 2010

### D H

Staff Emeritus
Re: Integrals

No, that's wrong. Alex dropped a factor of x.

The substitution $u=x^2$ does lead to $du=2x\,dx$. Solving for dx, $dx=1/(2x)\,du$, not $du/2$.

Alexx1: Try using this mixed form. (Alternately, look for a $2x\,dx$ in the integral.) using $$dx=1/(2\sqrt u)\,du$$ will just lead to confusion.

8. Jan 14, 2010

### Alexx1

Re: Integrals

I don't know how to find a 2x dx in the integral..
Can you explain it to me?

9. Jan 14, 2010

### Mentallic

Re: Integrals

What you have done wrong is that you failed to realize that the extra 2 in the answer you got can be attached to the constant of integration.
Take the derivative of both and you'll have the same result

10. Jan 14, 2010

### D H

Staff Emeritus
Re: Integrals

The integral in question is

$$\int x^3\,e^{x^2}\,dx$$

Rewrite this as

$$\int x^2\,e^{x^2}\,xdx$$

11. Jan 14, 2010

### Alexx1

Re: Integrals

Thanks!

12. Jan 14, 2010

### Alexx1

Re: Integrals

If du = 2xdx than xdx = du/2 ..

Than you get: (1/2) * integral x^2 e^u du..

Or am I wrong?

13. Jan 14, 2010

### D H

Staff Emeritus
Re: Integrals

Correct -- but incomplete. Why did you make the u-substitution in the exponential but not for the rest of integral?

14. Jan 14, 2010

### Alexx1

Re: Integrals

Ow you're right.. Stupid mistake.. Thank you very much, now I've solved it!

Can you check my other post please?

https://www.physicsforums.com/showthread.php?t=369488

15. Jan 14, 2010

### Alexx1

Re: Integrals

Can you also help me with this integral?

1/(1+cos(x)+sin(x)) dx

16. Jan 14, 2010

### Alexx1

Re: Integrals

Can you also help me with this integral?

1/(1+cos(x)+sin(x)) dx

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