# Integration by parts and substitution

## Homework Statement

Integrate: $$\sqrt{x}e^\sqrt{x}$$

See above

## The Attempt at a Solution

Well I started off first by taking t=sqrt(x) but that didn't get me very far. So then I decided to make x equal to t^2 which sort of worked. After hours of struggle I decided to have a look at the answers and I was pretty close but I am stumped by some steps in the answers.
Here's an image
[PLAIN]http://img689.imageshack.us/img689/8885/calcs.jpg [Broken]

First off I don't quite get how you are allowed to put t^2 in the differential.. no one ever taught me that. Then I don't understand how all of a sudden the first circled thing came about.

Does dt^2 mean that it's the second derivative? Did they take the integral in the second step (the circled part)? What happened there... Suddenly the square gets put on the t in the integrand and a 2 gets put outside of the integral.. whut?!
A friend of mine told me that dt^2 = dt*dt and then he blabbered on and got some weird solution...

Then the second circle. How the heck did they get the 2 there? It looks to me that they used t^2 as u and e^t as dv.

This is really mind blowing... because of this problem, everything I have learned in calculus seams useless because none of the rules make sense now to me... :(

Please help! :)

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## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
You have
$$\int \sqrt{x}e^{\sqrt{x}}dx$$
and make the substitution $t= \sqrt{x}$ which is the same as $x= t^2$.
$dx= 2t dt$ so the integral becomes
$$\int (t) e^{t} (2t dt)= 2\int t e^t dt[/itex]. "Putting $t^2$ inside the integral" is the same as making the substitution $x= t^2$ which I am sure you are familiar with. CompuChip Science Advisor Homework Helper I always get a bit confused by this notation as well. I learned it as follows: you are substituting x = t2. Then [tex]\frac{dx}{dt} = 2 t$$.
Now if you treat the derivative notation as an ordinary fraction (of course, this is technically not allowed, although it turns out that it works, and doing it the rigorous way would distract again from the point), you get
$$dx = 2 t \, dt$$.
This means that if you express x in t everywhere in the integrand
$$\int \sqrt{x} e^{\sqrt{x}} \, dx$$
you also have to express the dx in dt.
Thus you get
$$\int t e^{t} \cdot 2 t \, dt$$.

Some people prefer first replacing x by t2 in "dx", which gives d(t2) or, if you're lazy and/or don't like brackets, just d(t2).

For the second circle, the 2 is just the constant factor in front of the integral.
Obviously, if integration by parts gives
$$\int t^2 e^t \, dt = t^2 e^t - \int 2 t e^t \, dt$$
then it also gives
$$2 \int t^2 e^t \, dt = 2 \left( t^2 e^t - \int 2 t e^t \, dt \right)$$.

Dick
Science Advisor
Homework Helper
The notation d(f(t)) means f'(t)*dt. So d(t^2)=2*t*dt. That's where the '2' comes from. It's the rule you use anytime you do something like a u-substitution. Similarly they are also using d(e^t)=e^t*dt. What they are doing at the second red circle is an integration by parts on t^2*d(e^t). And BTW your original substitution sqrt(x)=t is the same substitution they are using. It should give the same answer if you do it right.

Thanks so much guys! I actually understand it now! I'm still working on the '2' and the '4' where they are integrating by parts...what's the rule for that when a constant is outside of the integral? Do you just multiply the entire integration by parts step by that constant? MY book doesn't make it any clearer either (caclulus early transcendentals)

Dick
Science Advisor
Homework Helper
Thanks so much guys! I actually understand it now! I'm still working on the '2' and the '4' where they are integrating by parts...what's the rule for that when a constant is outside of the integral? Do you just multiply the entire integration by parts step by that constant? MY book doesn't make it any clearer either (caclulus early transcendentals)

Sure. It doesn't matter if you write a multiplicative constant inside or outside the integral. They are both the same.

yeah i actually figured it out by just doing it on paper... This was actually pretty easy once I was done =] Go calculus :D