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Multiple problems to solve by means of integration

  1. Oct 24, 2006 #1
    Recently I have been given multiple problems to solve by means of integration. There are some problems which I am unsure on how to go about solving, or just don’t know. All help is greatly appreciated.


    Find the indefinite integral of [tex]\int {xe^{x^2 } } \;dx[/tex]

    This problem I simply do not know how to solve.


    Find the indefinite integral of [tex]\int {\frac{7}{{x^2 + 4x + 12}}} \;dx[/tex]

    This problem I began by trying to find factors of denominator of the fraction however found none. My reasoning on how to go about solving this problem is shown below, however I am unsure about my final answer.

    f\left( x \right) = \int {\frac{7}{{x^2 + 4x + 12}}} \;dx \\
    \int {\frac{7}{{\left( {x + 2} \right)^2 + 8}}} \;dx \\
    \int {\frac{7}{{u^2 + 8}}} \;dx\quad \quad u = x + 2 \\
    u' = \frac{{du}}{{dx}} \\
    dx = \frac{{du}}{{u'}} \\
    dx = \frac{{du}}{1} \\
    \int {\frac{7}{{u^2 + 8}}} \;du \\
    7\int {\frac{1}{{u^2 + 8}}\,du} \\
    f\left( x \right) = \ln \left| {u^2 + 8x} \right| + c \\
    f\left( x \right) = \ln \left| {\left( {x + 2} \right)^2 + 8x} \right| + c \\
    f\left( x \right) = \ln \left| {x^2 + 12x + 4} \right| + c \\


    There are others, but I'll only post these for now. I'll go back and try the others again and I may not need to post them here. Thanks in advance again,
  2. jcsd
  3. Oct 24, 2006 #2
    For the first do the sub. u = x^2

    As for the second problem, the integral of 1/(u^+8) is not ln(u^+8). One way to check if you've got the right answer, is to take the derivative of the result and see if you obtain the integrand.

    int(1/(u^+8)) is quite easy. Do you know the integral of 1/(x^2+a^2), where 'a' is a constant?
  4. Oct 24, 2006 #3
    I am unfamiliar with the integral of 1/(x^2+a^2), where 'a' is a constant, could you please explain how to derive this integral. Thanks.
  5. Oct 24, 2006 #4


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    Homework Helper

    It is a 'table integral', and it equals [tex]\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan (\frac{x}{a})+C[/tex].
  6. Oct 24, 2006 #5
    For that use the sub. [tex]x = a\tan{\theta}[/tex].
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