Maths & Physics - Solved Questions

Monday, August 11, 2008

Q) Let f(x) = $\int \frac{{x}^{2}}{( 1 + {x}^{2}) ( 1 + \sqrt{1 + {x}^{2)}} }dx$ and f(0) = 0 . Then f(1) = ?

A)

$\int \frac{{x}^{2}}{( 1 + {x}^{2}) ( 1 + \sqrt{1 + {x}^{2)}} }dx$

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

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

By Applying partial fractions -

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

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

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

= $- tan^{-1}x + log ( x + \sqrt{1 + x^2}) + C$

BY using the statement that f(0) = 0 ........We get C = 0

$f(1) = - tan^{-1}(1) + log (1 + \sqrt{2})$

$f(1) = log (1 + \sqrt{2}) - \frac{\Pi}{4}$

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Monday, August 11, 2008