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  1. #1
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    integral , basic ..

    Use this formula only $\displaystyle \int x^n ~dx = \frac{ x^{n+1}}{n+1} + c$

    Find the integral $\displaystyle \int \frac{dx}{\sqrt{ x + \sqrt{ x^2 + 1}}}$


    Of course , little Algebra is allowed . Try to answer this without applying substitution , integration by parts , etc
    Last edited by simplependulum; Oct 31st 2009 at 06:06 AM.
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  2. #2
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    Edit
    Last edited by Deadstar; Nov 9th 2009 at 04:23 PM.
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  3. #3
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    What is the solution
    i lost 2 hours from my life
    and BTW, you should said that n dont equal -1
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  4. #4
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    intermediate step

    I'm considering squaring.
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  5. #5
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    It seems impossible to solve without using substitution
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  6. #6
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    Below is the solution to this problem ,


    Spoiler:


    To avoid appearing the imaginary number $\displaystyle i $ in the solution ,

    Consider

    $\displaystyle \int \frac{dx}{\sqrt{ x + \sqrt{ x^2 - a^2 }}}$

    $\displaystyle = \int \frac{dx}{\sqrt{ x + \sqrt{ x^2 + 1}}} \cdot \frac{ \sqrt{ x - \sqrt{x^2 - a^2 }} }{\sqrt{ x - \sqrt{x^2 - a^2 }}}$

    $\displaystyle = \int \frac{ \sqrt{ x - \sqrt{x^2 - a^2 }} }{ \sqrt{ x^2 - (x^2 - a^2)}}~dx $

    $\displaystyle = \frac{1}{a} \int \sqrt{ x - \sqrt{x^2 - a^2 }} ~dx $

    Now , the integral looks much more simplier but that's not enough , we have to do something special in the following steps :

    $\displaystyle = \frac{1}{a} \int \sqrt{\frac{2( x - \sqrt{x^2 - a^2})}{2}} ~dx $

    $\displaystyle = \frac{1}{a \sqrt{2}} \int \sqrt{2x - 2\sqrt{(x-a)(x+a)}}~dx $

    $\displaystyle = \frac{1}{a \sqrt{2}} \int \sqrt{ ( x+a) - 2\sqrt{(x-a)(x+a)} + (x-a)}~dx $

    $\displaystyle = \frac{1}{a \sqrt{2}} \int \sqrt{ \sqrt{x+a}^2 - 2\sqrt{(x-a)(x+a)} + \sqrt{x-a}^2}~dx $

    $\displaystyle = \frac{1}{a \sqrt{2}} \int \sqrt{ ( \sqrt{x+a} - \sqrt{x-a})^2}~dx $

    $\displaystyle = \frac{1}{a \sqrt{2}} \int [\sqrt{x+a} - \sqrt{x-a} ]~dx$

    $\displaystyle = \frac{\sqrt{2}}{3a} \left[ \sqrt{x+a}^3 - \sqrt{x-a}^3 \right ] + C $

    $\displaystyle = \frac{\sqrt{2}}{3a} ( \sqrt{x+a} - \sqrt{x-a})( (x+a) + \sqrt{x^2 - a^2} + (x-a) ) + C $

    $\displaystyle = \frac{2}{3a} \sqrt{ x - \sqrt{x^2 - a^2}} ( 2x + \sqrt{ x^2 - a^2}) + C $

    $\displaystyle = \frac{2}{3} \frac{1}{\sqrt{x+ \sqrt{x^2 - a^2}}} ( 2x + \sqrt{x^2 - a^2} ) + C $

    Sub. $\displaystyle -a^2 = +1 $

    the integral

    $\displaystyle = \frac{2}{3} \frac{2x + \sqrt{x^2 +1} }{\sqrt{x+ \sqrt{x^2 +1}}} + C $

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  7. #7
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    Quote Originally Posted by TWiX View Post
    What is the solution
    i lost 2 hours from my life
    and BTW, you should said that n dont equal -1


    I just forgot to add this point but actually it is also true when $\displaystyle n = -1 $

    Although , from the formula , it gives us something like this :

    $\displaystyle \frac{ x^0}{0} + C $ or $\displaystyle \lim_{a\to 0} \frac{x^a}{a} + C $

    the numerator tends to $\displaystyle 1$ while the denominator tends to zero .

    ,if we rewrite the constant $\displaystyle C $ as $\displaystyle -\frac{1}{a} + C'$ , and make $\displaystyle a $ very small $\displaystyle \to 0 $

    so we have

    $\displaystyle \int \frac{dx}{x} = \lim_{a\to 0} \frac{x^a -1 }{a} + C' $

    Then use L rule to evaluate the limit , we will find that

    $\displaystyle I = \ln(x) + C'$
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