A) (2, 3)
B) (2, 1)
C) \[(-2,1)\]
D) \[(-2,3)\]
Correct Answer: A
Solution :
Let\[\tan x=t\Rightarrow dt={{\sec }^{2}}xdx\] \[=(1+{{\tan }^{2}}x)dx\] \[=(1+{{t}^{2}})dx\] \[\Rightarrow \]\[dx=\frac{dt}{1+{{t}^{2}}}\] \[\therefore \] \[\int_{{}}^{{}}{\frac{\tan x}{{{\tan }^{2}}x+\tan x+1}dx}\] \[=\int_{{}}^{{}}{\frac{t}{(1+{{t}^{2}})(1+t+{{t}^{2}})}dt}\] \[={{\tan }^{-1}}t-\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right)\] \[=x-\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan x+1}{\sqrt{3}} \right)\] Clearly then \[(K,A)=(2,3).\]so option A is the correct answer.\[\int_{{}}^{{}}{\frac{\tan x+1+{{\tan }^{2}}x}{\tan x+1+{{\tan }^{2}}x}}dx-\int_{{}}^{{}}{\frac{(1+{{\tan }^{2}}x)}{1+\tan +{{\tan }^{2}}x}}dx\] \[=x-\int_{{}}^{{}}{\frac{{{\sec }^{2}}dx}{1+\tan x+{{\tan }^{2}}x}}\] \[=x-\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+t+\frac{1}{4}+1-\frac{1}{4}}}\] \[=x-\int_{{}}^{{}}{\frac{dt}{{{\left( t+\frac{1}{2} \right)}^{2}}+{{\left( \frac{3}{2} \right)}^{2}}}}\] \[=x-\frac{2}{{{c}_{3}}}{{\tan }^{-}}\left( \frac{t+{{11}_{2}}}{{{c}_{3}}/2} \right)+c\] \[x-\frac{2}{{{c}_{3}}}{{\tan }^{-}}\left( \frac{2\tan x+1}{\sqrt{3}} \right)+c\] \[A=3\] \[K=2.\] Thus the answer is (2, 3)
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