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If $\int \frac{x}{x^{2} - 4 x + 8} d x = K \log (x^{2} - 4 x + 8) +$ $\tan^{- 1} (\frac{x - 2}{2}) + C$ then the value of $K$ is

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Completing the square, we getx2−4x+8=(x−2)2+4 so put x−2=u∫xx2−4x+8dx=∫u+2u2+4du=12∫2uu2+4du+2∫duu2+4=12log⁡u2+4+tan−1⁡u2+C=12log⁡x2−4x+8+tan−1⁡x−22+C.

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If ∫xx2−4x+8dx=Klog⁡x2−4x+8+tan−1⁡x−22+C then the value of K is

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If $\int \frac{x}{x^{2} - 4 x + 8} d x = K \log (x^{2} - 4 x + 8) +$ $\tan^{- 1} (\frac{x - 2}{2}) + C$ then the value of $K$ is