# Integration question

1. Jul 1, 2016

### chwala

[2][/2]1. The problem statement, all variables and given/known data
$∫e^x^2 + 2e^x^2x^2 dx$

2. Relevant equations

3. The attempt at a solution
i let$u= x^2, ⇒ du = 2x dx, ⇒∫e^x^2 dx = e^x^2/2x$ is this correct? by using integration by parts..... i am getting $xe^x^2-e^x^2/2x$

2. Jul 1, 2016

### Math_QED

You can verify whether you are correct by using wolframalpha.com. Also, your code has an error.

3. Jul 1, 2016

### Abhishek Sethi

You may use integration by parts on the first term.
$$\int e^{x^2} \, dx=xe^{x^2}-\int x(2xe^{x^2}), dx$$
The equation above is sufficient for your problem.
Hope it helps.

4. Jul 1, 2016

### chwala

i can see you are using integration by parts, what is your $u, du, v, dv$?

5. Jul 1, 2016

### Abhishek Sethi

I am not sure what you exactly mean by $u,v,du,dv$. To put it more clearly,
$$\int e^{x^2} \, dx=(e^{x^2})\int 1\ dx-\int x\frac{d(e^{x^2})}{dx} dx$$

6. Jul 1, 2016

### chwala

this is something "new " to me in my understanding integration by parts is $∫udv= uv - ∫vdu$ unless you have used a different approach, your method can't be described as integration by parts as stated in your opening statement.

7. Jul 1, 2016

### chwala

i can see you picked $u= e^x^2, dv=1 ⇒v=x, u'=2xe^x^2$

8. Jul 1, 2016

### chwala

Thanks Abishek...........greetings from Africa

9. Jul 1, 2016

### chwala

i dont know when you continue with the integration of the second term the function is getting complicated i.e if you ∫2x^2e^x^2 dx or are we picking again
dv = 1, u= 2x^2e^x^2?

10. Jul 1, 2016

### chwala

on ∫x.2xe^x^2 dx i let dv= x^2⇒ v=(x^3)/3, u= e^x^2 ⇒ du = 2x.e^x^2 thus
∫x.2xe^x^2 dx = (2e^x^2.x^3)/3-2 ∫(x^3)/3.2x.e^x^2 dx without simplification...this to me gets more complicated and no final answer is possible...

11. Jul 1, 2016

### chwala

lol lol lol i was blind but now i can see...

∫(e^x^2+2e^x^2x^2)dx= xe^x^2- ∫2e^x^2x^2dx + ∫2e^x^2x^2dx + k
= xe^x^2 + k

chikhabi from Africa lol

12. Jul 1, 2016

### SammyS

Staff Emeritus
It's good to see you trying to use LaTeX, but you are having some problems.

Is this what you mean? $\displaystyle\ \int \left( e^{x^2} + 2e^{x^2}x^2 \right) dx\$ It's difficult to tell with all of those run-on exponents.

13. Jul 2, 2016

### chwala

yes that is what i mean...how do i type that in latex?

14. Jul 2, 2016

### SammyS

Staff Emeritus

Otherwise, for some LaTeX here in PF that you don't such access to: Right click on the LaTeX expression. Click on "Show Math As". Click on "TeX Commands". A pop-up window should appear with the LaTeX code.

The \displaystyle LaTeX command you see in my code gives larger format when doing "in-line" LaTeX using the $\ \text{$ ... $}\$ delimiters.

Also, look up LaTeX tutorial here in Physics Forums.

15. Jul 3, 2016

### chwala

$\left| \frac a b \right|$ am trying practise on latex.......
$\left[ 1 - \left( \frac a x \right)^2 \right]^{-1/3}$
$\int x^2e^x \, dx$
${ \displaystyle\ \int \left( e^{x^2} + 2e^{x^2}x^2 \right) dx\ }$​

Last edited: Jul 4, 2016
16. Jul 4, 2016

### Ray Vickson

You can also do it in plain text, just by using parentheses correctly. When you write e^x^2 that could mean $e^{x^2}$ or $(e^x)^2 = e^{2x}$. In plain text you would write the first as e^(x^2) and the second as (e^x)^2.