∫Tan−1⁡8x−43−4x2+4xdx=

∫tan−1⁡8x−43−4x2+4xdx=∫tan−1⁡4(2x−1)4−(2x−1)2dx=∫tan−1⁡(2x−1)1−2x−122dx=∫2tan−1⁡(2x−1)2dx 2x-12=t⇒dx =dt =2∫tan-1tdt=2t tan-1t-∫2t1+t2dt=22x-12 tan-12x-12-log1+2x-122+C =(2x-1) tan-12x-12-log4x2-4x+5+C

∫Tan−1⁡8x−43−4x2+4xdx=

∫tan−1⁡8x−43−4x2+4xdx=∫tan−1⁡4(2x−1)4−(2x−1)2dx=∫tan−1⁡(2x−1)1−2x−122dx=∫2tan−1⁡(2x−1)2dx 2x-12=t⇒dx =dt =2∫tan-1tdt=2t tan-1t-∫2t1+t2dt=22x-12 tan-12x-12-log1+2x-122+C =(2x-1) tan-12x-12-log4x2-4x+5+C

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$\int {Tan}^{- 1} (\frac{8 x - 4}{3 - 4 x^{2} + 4 x}) d x =$

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$(2 x - 1) {Tan}^{- 1} (\frac{2 x - 1}{2}) - \log (4 x^{2} + 4 x + 5) C$

$(2 x - 1) {Tan}^{- 1} (\frac{2 x - 1}{2}) - \frac{1}{2} \log (4 x^{2} - 4 x - 5) + C$

$(2 x + 1) {Tan}^{- 1} (\frac{2 x - 1}{2}) - \log \sqrt{(2 x + 1)^{2} + 1} + C$

$(2 x - 1) {Tan}^{- 1} (\frac{2 x - 1}{2}) - \log (4 x^{2} - 4 x + 5) + C$

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Detailed Solution

$\begin{array}{l} \int \tan^{- 1} (\frac{8 x - 4}{3 - 4 x^{2} + 4 x}) d x = \int \tan^{- 1} (\frac{4 (2 x - 1)}{4 - (2 x - 1)^{2}}) d x \\ = \int \tan^{- 1} (\frac{(2 x - 1)}{1 - {(\frac{2 x - 1}{2})}^{2}}) d x \\ = \int 2 \tan^{- 1} (\frac{(2 x - 1)}{2}) d x (\frac{2 x - 1}{2}) = t \Rightarrow d x = d t = 2 \int \tan^{- 1} t d t = 2 t \tan^{- 1} t - \int \frac{2 t}{1 + t^{2}} d t \\ = 2 (\frac{2 x - 1}{2}) \tan^{- 1} (\frac{2 x - 1}{2}) - \log (1 + {(\frac{2 x - 1}{2})}^{2}) + C = (2 x - 1) \tan^{- 1} (\frac{2 x - 1}{2}) - \log (4 x^{2} - 4 x + 5) + C \end{array}$