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Ak2ng
Ak2ng asked in Science & MathematicsMathematics · 1 decade ago

Integration of Function??

Was wondering if anyone knew how to integrate

(10*r^2)/(r^2+a^2) where a= some constant.

Thanks!

4 Answers

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  • Joe L
    Lv 5
    1 decade ago
    Favorite Answer

    If you don't mind, I'm going to replace your "r" by "x"

    so you have

    ∫ (10a²) dx/(x² - a²)

    The first thing to do is a synthetic division. Divide the denominator into the numerator. This gives you:

    ∫[10 - 10a² dx /(x² + a²)]

    which is

    10x - 10a²∫dx/(x² + a²)

    That last term is a standard integral, so the net result is:

    10x - 10a²[1/a(arctan(x/a)) + C =

    10x - 10a[arctan(x/a)] + C =

    10[x - a(arctan(x/a))] + C

  • yasser t
    Lv 5
    1 decade ago

    10r^2/(r^2+a^2)= 10-(10a^2)/(r^2+a^2) =

    10- 10/((r/a)^2+1) the integral = 10r- 10a*arctan(r/a)+c

  • santmann2002
    Lv 7
    1 decade ago

    = 10 * (r^2+a^2 -a^2)/ r^2+a^2) = 10 -10a^2 * 1/(r^2+a^2) =

    10r -10 *1/(1+r^2/a^2) =10(r +arctg(r/a)) *a +C

  • mister math
    1 decade ago

    you use integration by parts i think

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