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

  1. Mar 21, 2006 #1
    Hello, I have a few questions! I need clarification on certain points that were not very clear in my calculus book.

    Question 1:

    I know that [tex] \int e^{ax} dx = \frac{1}{a} e^{ax} [/tex]

    But how do you integrate [tex] \int e^{ax^2} dx [/tex]?

    Question 2:

    I know that integrating by parts is [tex] \int (something) dx= uv - \int vdu [/tex]

    But what if there is a range?

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

    Question 3:

    How do you integrate [tex] \int log_ax dx [/tex] and [tex] \int e^{ln|secx|} dx [/tex].

    In fact, is [tex] e^{ln|secx|} [/tex] reducable?

    Question 4:

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

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

    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?

    Reply to HallsofIvy:

    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 [tex] \int xe^{x^2/2} [/tex] and the answer is [tex] e^{x^2/2} + C [/tex] but I don't understand how to do it!
    Last edited: Mar 22, 2006
  2. jcsd
  3. Mar 21, 2006 #2


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    Answer to question 1: You don't. That integral,
    [tex]\int e^{ax^2}dx[/tex]
    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
    [tex]/int xe^{ax^2}dx[/tex]
    then you could make the substitution [itex]u= e^{ax^2}[/itex] 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
  4. Mar 22, 2006 #3
    Notice that [tex]\frac{d}{dx}\left(\frac{x^2}{2}\right)=x[/tex]
    So it is actually [tex]\int{f'\left(x\right)e^{f\left(x\right)}=e^{f\left(x\right)}+c[/tex]
  5. Mar 26, 2006 #4
    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)

    Question 6

    How do you integrate:

    [tex] \int \frac{1}{x^a} dx [/tex]

    [tex] \int \frac{1}{a^x} dx [/tex]

    [tex] \int \frac{1}{a^{bx}} dx [/tex]

    [tex] \int \frac{1}{a^{bx^{c}}} dx [/tex]

    Should I have learnt this somewhere? I don't see these in the integral tables or rules..
    Last edited: Mar 26, 2006
  6. Mar 26, 2006 #5


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    Science Advisor

    [tex]\frac{1}{x^a}= x^{-a}[/tex]
    Use the power rule.

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

    Same thing:
    [tex]\frac{1}{a^{bx}}= a^{-bx}[/tex]
    Make the substitution u= -bx.

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