# Integration question

Gold Member
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## Homework Statement

##∫e^x^2 + 2e^x^2x^2 dx##[/B]

## 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##

member 587159

## Homework Statement

##∫e^x^2 + 2e^x^2x^2 dx##[/B]

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

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.

Gold Member
You may use integration by parts on the first term of the equation.
$$\int e^{x^2} \, dx=xe^{x^2}-\int x(2xe^{x^2}), dx$$
The equation above is sufficient for your problem.
Hope it helps.
i can see you are using integration by parts, what is your ##u, du, v, dv##?

i can see you are using integration by parts, what is your ##u, du, v, dv##?
I am not sure what you exactly mean by ##u,v,du,dv##. To put it more clearly,
$$\int e^{x^2} \, dx=(e^{x^2})\int 1\ dx-\int x\frac{d(e^{x^2})}{dx} dx$$

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

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

Gold Member
Thanks Abishek...........greetings from Africa

Gold Member
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?

Gold Member
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...

Gold Member
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

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

## ∫ e^x^2 + 2e^x^2x^2 dx##
##\int e^{x^2} + 2e^{x^2}x^2 dx## (fixed by SammyS.)

## 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##
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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
yes that is what i mean...how do i type that in latex?

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.

Gold Member
##\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:
Ray Vickson
Homework Helper
Dearly Missed
yes that is what i mean...how do i type that in latex?
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.

chwala