How Can I Solve These Complex Integrals?

[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..

 

[tiny-tim]

Hi Alexx1!

(have an integral: ∫ and a square-root: √ and try using the X² tag just above the Reply box )

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..

For 1 and 2, use the obvious substitutions.

 

[Mentallic]

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!

 

[Alexx1]

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!

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)

 

[HallsofIvy]

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

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

u = x^2 --> du = 2xdx --> dx = du/2

Yes, that's right.

?

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

 

[Alexx1]

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


Yes, that's right.

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?

 

[D H]

u = x^2 --> du = 2xdx --> dx = du/2

Yes, that's right.

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.

 

[Alexx1]

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


Yes, that's right.

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.

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

 

[Mentallic]

= 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?

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

 

[D H]

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

The integral in question is

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

Rewrite this as

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

 

[Alexx1]

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

Thanks!

 

[Alexx1]

The integral in question is

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

Rewrite this as

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

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

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

Or am I wrong?

 

[D H]

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

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

Or am I wrong?

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

 

[Alexx1]

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

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

 

[Alexx1]

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

Can you also help me with this integral?

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

 

[Alexx1]

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

Can you also help me with this integral?

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