# If $\int \frac{\tan x}{1+\tan x +\tan^{2}x}dx=$$x- \frac{K}{\sqrt{A}}\tan ^{-1} (\frac{K\tan x+1} {\sqrt{A}})+c$(C is a constant of integration), then the ordered pair (K, A) is equal to : Option 1) (2, 1) Option 2) (−2, 3) Option 3) (2, 3) Option 4) (−2, 1)

A Anuj Pandey
H Himanshu

As we have learned

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

 Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable ,$\left ( t\, or\ \theta \right )$

Integration by perfect square method -

$ax^{2}+bx+c=a\left [ x^{2}+\frac{bx}{a}+\frac{c}{a} \right ]=a\left [ \left ( x+\frac{b}{2a} \right )^{2}+\frac{c}{a}-\frac{b^{2}}{4a^{2}} \right ]$

- wherein

 Make the coefficient of $x^{2}$ +ve one .

put   $\tan x = t ; \sec^{2}x = dt$

$\int \frac{t}{(1+t+t^{2})(1+t^{2})}dt= \int \frac{1}{1+t^{2}}- \frac{1}{(1+t+t^{2})}dt$

$\tan ^{-1}(t)- \int \frac{dt}{(t+1/2)^{2}+ (\sqrt3/2)^{2}}$

$\tan ^{-1}(t)- \frac{1}{\sqrt3/2} \tan \frac{2t+1}{\sqrt3}$

thus k= 2  and A= 3

Option 1)

(2, 1)

This is incorrect

Option 2)

(−2, 3)

This is incorrect

Option 3)

(2, 3)

This is correct

Option 4)

(−2, 1)

This is incorrect

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