If-tan-x-1-tan-x-tan-2-x-dx-x-k-A-tan-1-k-tan-x-1-A-C-where-C-is-constant-of-integration-then-the-ordered-pair-k-A-is-equal-to-

Question Number 133004 by liberty last updated on 18/Feb/21

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

where C is constant of integration .

then the ordered pair (k, A) is

equal to

Answered by EDWIN88 last updated on 18/Feb/21

consider \tan x = 1 + \tan^{2} x + \tan x - \sec^{2} x

I = \int \frac{1 + \tan^{2} x + \tan x - \sec^{2} x}{1 + \tan x + \tan^{2} x} dx

I = \int 1 - \frac{\sec^{2} x}{1 + \tan x + \tan^{2} x} dx

I = x - \int \frac{\sec^{2} x}{1 + \tan x + \tan^{2} x} dx

put u = \tan x

I = x - \int \frac{du}{1 + u + u^{2}} = x - \int \frac{du}{{(u + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}

I = x - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{u + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C

I = x - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + C

we find \to {\begin{cases} k = 2 \\ A = 3 \end{cases} \Rightarrow (k, A) = (2, 3)

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