A) \[\tan 1-\frac{\pi }{4}\]
B) \[\tan 1+1\]
C) \[\frac{\pi }{4}\]
D) \[1-\frac{\pi }{4}\]
Correct Answer: A
Solution :
Let\[\int_{{}}^{{}}{\left( \frac{{{x}^{2}}+{{\sin }^{2}}x}{1+{{x}^{2}}} \right){{\sec }^{2}}xdx}\] \[=\int_{{}}^{{}}{\frac{{{x}^{2}}{{\sec }^{2}}+\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{1+{{x}^{2}}}dx}\] \[=\int_{{}}^{{}}{\frac{{{x}^{2}}{{\sec }^{2}}x+{{\tan }^{2}}x}{1+{{x}^{2}}}dx}\] \[=\int_{{}}^{{{x}^{2}}}{\frac{{{x}^{2}}\left( 1+{{\tan }^{2}}x \right)+{{\tan }^{2}}x}{1+{{x}^{2}}}dx}\] \[=\int_{{}}^{{}}{\frac{{{x}^{2}}+{{\tan }^{2}}x(1+{{x}^{2}})}{1+{{x}^{2}}}dx}\] \[=\int_{{}}^{{}}{\frac{{{x}^{2}}}{1+{{x}^{2}}}dx}+\int_{{}}^{{}}{{{\tan }^{2}}xdx}\] \[=\int_{{}}^{{}}{\frac{{{x}^{2}}+1-1}{1+{{x}^{2}}}dx}+\int_{{}}^{{}}{\left( {{\sec }^{2}}x-1 \right)}dx\] \[=\int_{{}}^{{}}{1dx-}\int_{{}}^{{}}{\frac{dx}{1+{{x}^{2}}}}x\int_{{}}^{{}}{{{\sec }^{2}}xdx-\int_{{}}^{{}}{dx}}\] \[=-{{\tan }^{-1}}x+\tan +c\] Given:\[f(0)=0\] \[\Rightarrow \]\[f(0)=-{{\tan }^{-1}}0+\tan 0+c\]\[\Rightarrow \]\[c=0\] \[\therefore \]\[f(x)=-{{\tan }^{-1}}x+\tan x\]Now, \[f(1)=-{{\tan }^{-1}}(1)+\tan 1=\tan 1-\frac{\pi }{4}\]
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