The integral of value of x such that that Tan−1xx−2−Tan−1x2x−1=Tan−123is ............

Tan−1xx−2−Tan−1x2x−1=Tan−123⇒tanTan−1xx−2−Tan−1x2x−1=23⇒xx−2−x2x−11+x(x−2)x2x−1=23⇒2x2−x−x2+2x2x2−5x+2+x2=23⇒3(x2+x)=2(3x2−4x+2)⇒3x2−11x+4=0⇒(x−4)(3x−1)=0⇒x=4 or 1/3.

The integral of value of x such that that Tan−1xx−2−Tan−1x2x−1=Tan−123is ............

Tan−1xx−2−Tan−1x2x−1=Tan−123⇒tanTan−1xx−2−Tan−1x2x−1=23⇒xx−2−x2x−11+x(x−2)x2x−1=23⇒2x2−x−x2+2x2x2−5x+2+x2=23⇒3(x2+x)=2(3x2−4x+2)⇒3x2−11x+4=0⇒(x−4)(3x−1)=0⇒x=4 or 1/3.

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$T h e int e g r a l o f v a l u e o f x s u c h t h a t t h a t T a n^{- 1} (\frac{x}{x - 2}) - T a n^{- 1} (\frac{x}{2 x - 1}) = T a n^{- 1} (\frac{2}{3}) i s ............$

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Detailed Solution

$\begin{array}{l} T a n^{- 1} \frac{x}{x - 2} - T a n^{- 1} \frac{x}{2 x - 1} = T a n^{- 1} \frac{2}{3} \\ \Rightarrow \tan [T a n^{- 1} \frac{x}{x - 2} - T a n^{- 1} \frac{x}{2 x - 1}] = \frac{2}{3} \Rightarrow \frac{\frac{x}{x - 2} - \frac{x}{2 x - 1}}{1 + \frac{x}{(x - 2)} \frac{x}{2 x - 1}} = \frac{2}{3} \Rightarrow \frac{2 x^{2} - x - x^{2} + 2 x}{2 x^{2} - 5 x + 2 + x^{2}} = \frac{2}{3} \\ \Rightarrow 3 (x^{2} + x) = 2 (3 x^{2} - 4 x + 2) \Rightarrow 3 x^{2} - 11 x + 4 = 0 \Rightarrow (x - 4) (3 x - 1) = 0 \Rightarrow x = 4 o r 1 / 3. \end{array}$