01-08-10 Integration....

Q→1

I = \int \sqrt{x + \sqrt{x^{2} + a^{2}}}

{x + \sqrt{x^{2}+ a^{2}}} = t → eq1

\left(1 + x/\sqrt{x^{2} + a ^{2}} \right)dx = dt

\left[ \left(\sqrt{x^{2}+ a^{2}} + x \right)/\left(\sqrt{x^{2} + a^{2}} \right)\right]dx = dt →eq3

also by rationalisation you can see that ,

-a²t = \left(x - \sqrt{x^{2} + a^{2}} \right) → eq2

subtracting eq1 - eq2 we get,

t + a²t = 2 \sqrt{x^{2} + a^{2}}

ie \sqrt{x^{2} + a^{2}} = \frac{t^{2} + a^{2}}{2t}

putting this value of \sqrt{x^{2} + a^{2}} in eq3 we get,
(\frac{2t^{2}}{a^{2} + t^{2}})dx = dt

therefore,
I = \int t^\frac{1}{2}( \frac{1}{2} + \frac{a^{2}}{2t^{2}})dt

I = \frac{t^{\frac{3}{2}}}{3} - \frac{a^{2}}{t^{\frac{1}{2}}}

(we know the value of t from eq 1 )

PLZ DO TELL ANY MISTAKES IF ANY

i think there can be another smaller solution by putting x= atanθ
but im not able to proceed after a certain step in this method plz someone try this method also

Vaibhav ... ur absolutely right ...

I think ur stuck in the step where the expr. is ....

∫√(asecθ + atanθ) asec²Î¸ dθ ....

Amn't I right ?????!!!!!

Let .............. x + √x ² + a ² = z ²

( x - z ² ) ² = x ² + a ² ;

or , x ² - 2 z ² x + z ⁴ = x ²+ a ²

or , 2 z ² x = z ⁴ - a ²

or , x = z ²2 - a ²2 z ²

or , dx = dz { z2 + a ²z ³ }

Now , I = ∫ z { z2 + a ²z ³ } dz

Which is not troublesome , I think ?