# Integrating Exponentials with Roots that have Roots? (And other small Q's)

1. Mar 21, 2006

### kape

Hello, I have a few questions! I need clarification on certain points that were not very clear in my calculus book.

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Question 1:

I know that $$\int e^{ax} dx = \frac{1}{a} e^{ax}$$

But how do you integrate $$\int e^{ax^2} dx$$?

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Question 2:

I know that integrating by parts is $$\int (something) dx= uv - \int vdu$$

But what if there is a range?

If it is $$\int_{a}^{b} (something) dx$$ does it equal $$\left[ uv \right]_{a}^{b} - \int_{a}^{b} vdu$$ or does it simply equal $$uv - \int_{a}^{b} vdu$$?

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Question 3:

How do you integrate $$\int log_ax dx$$ and $$\int e^{ln|secx|} dx$$.

In fact, is $$e^{ln|secx|}$$ reducable?

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Question 4:

I was taught that arcsinx exist only in the range $$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$$ and $$\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$$ (I think because it fails the horizontal test if it isn't in those ranges)

If so, is it possible to integrate $$\int_{0}^{\pi} xarcsinx dx$$? (If it is possible, is it because it isn't simply arcsinx but xarcsinx?)

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Question 5:

I am having a lot of problems integrating fuctions with exponents etc that have complex roots. My elementary calculus is shaky at best and I'm taking Advanced Engineering Mathematics (Kreyzig) - I have to. Can anyone recommend me any links or books that may help me?

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Thank you for your reply! I have a question about your reply on question 1: In my Adv Eng Maths (Kreyzig) book, one of the questions is how to integrate $$\int xe^{x^2/2}$$ and the answer is $$e^{x^2/2} + C$$ but I don't understand how to do it!

Last edited: Mar 22, 2006
2. Mar 21, 2006

### HallsofIvy

Answer to question 1: You don't. That integral,
$$\int e^{ax^2}dx$$
is well known not to have an elementary integral. In fact, precisely that integral (with a= -1) is important in Statistics and it's integral is defined to be "Erf(x)", the error function.

If it were
$$/int xe^{ax^2}dx$$
then you could make the substitution $u= e^{ax^2}$ and have the xdx already for du= 2xdx.

Question 2, Yes, just plug the limits of integration into the formula.

Question 3 seems to have disappeared.

Last edited by a moderator: Mar 22, 2006
3. Mar 22, 2006

### whkoh

Notice that $$\frac{d}{dx}\left(\frac{x^2}{2}\right)=x$$
So it is actually $$\int{f'\left(x\right)e^{f\left(x\right)}=e^{f\left(x\right)}+c$$

4. Mar 26, 2006

### kape

Thank you for answering questions 1 & 2.. I think I understand.

Sorry question 3 was deleted, don't quite know how that happened.

Also, I have one more question: (which is kind of similar to question 3)

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Question 6

How do you integrate:

$$\int \frac{1}{x^a} dx$$

$$\int \frac{1}{a^x} dx$$

$$\int \frac{1}{a^{bx}} dx$$

$$\int \frac{1}{a^{bx^{c}}} dx$$

Should I have learnt this somewhere? I don't see these in the integral tables or rules..

Last edited: Mar 26, 2006
5. Mar 26, 2006

### HallsofIvy

$$\frac{1}{x^a}= x^{-a}$$
Use the power rule.

$$\frac{1}{a^x}= a^{-x}$$
Make the substitution u= -x and then use
$$\int a^x dx= \frac{a^x}{ln a}$$

Same thing:
$$\frac{1}{a^{bx}}= a^{-bx}$$
Make the substitution u= -bx.

$$\int \frac{1}{a^{bx^c}}dx$$
depends strongly on what c is. There is no general anti-derivative.