IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

rpz

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

Then f(1) =

A ) log ( 1 + $\sqrt{2}$ )

B) log ( 1 + $\sqrt{2}$ ) + $\frac{\Pi}{4}$

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

D) None of these.

spideyunlimited

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

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

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

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

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

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

HENCE OPTION (C) is the RIGHT answer.

akhil_o

$Oterwise...putting \,x=tanA\\dx=sec^2AdA\\the\, integral \,becomes\\ \int \frac{tan^2A.dA}{1+secA}\\=\int \frac{sec^2A-1.dA}{secA+1}$

$\int \frac{(secA-1)(secA+1)dA}{secA+1}\\=\int secA-1dA$

$=ln|secA+tanA|-A+c\\=ln|x+\sqrt{1+x^2}|-tan^{-1}x+C$

and then proceed as spidey has done

spideyunlimited

ya that's also a good way.