Let $f(x) = \int_{}^{}\frac{x^{2}dx}{\left( 1 + x^{2} \right)(1 + \sqrt{1 + x^{2}})}$ and $f(0) = 0$. Then, $f(1)$ is

Question

Dr. Abubacker · Accepted Answer

(b)
On putting, $x = 	an	heta\  \Rightarrow \ \ dx = sce^{2}	heta\ d	heta$, we get
$f(x) = \int_{}^{}\frac{	an^{2}{	heta.\sec^{2}	heta}}{{\sec^{2}	heta}{(1 + \sec	heta)}}d	heta$
$= \int_{}^{}\frac{\sec^{2}{	heta - 1}}{1 + \sec	heta}d	heta = \int_{}^{}{(\sec	heta - 1)}d	heta$
$= \log{(\sec	heta + 	an	heta)} - 	heta + c$
$\Rightarrow \ \ f(x) = \log{(\sqrt{1 + x^{2}} + x)} - 	an^{- 1}x + c$
At $x = 0,\ \ \ f(0) = \log{(1 + 0) - 0 + c\ \  \Rightarrow \ \ c = 0\ }$
$	herefore\ \ f(x) = \log{(\sqrt{1 + x^{2}} + x)} - 	an^{- 1}x$
At $x = 1,\ \ f(1) = \log{(1 + \sqrt{2})} - \frac{\pi}{4}$

Q. Let $f(x) = \int_{}^{}\frac{x^{2}dx}{\left( 1 + x^{2} \right)(1 + \sqrt{1 + x^{2}})}$ and $f(0) = 0$. Then, $f(1)$ is

Explanation

Reviewed by Dr. Abubacker

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