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$\int \frac{\sqrt{x - 1}}{x \sqrt{x + 1}} d x is equal to$

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a

ln⁡x−x2−1−tan−1⁡x+C

b

ln⁡x+x2−1−tan−1⁡x+C

c

ln⁡x−x2−1−sec−1⁡x+C

d

ln⁡x+x2−1−sec−1⁡x+C

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detailed solution

Correct option is D

I=∫x−1xx+1dx=∫x−1xx2−1=∫dxx2−1−∫dxxx2−1=ln⁡x+x2−1−sec−1⁡x+C

Similar Questions

If $\int \frac{x + 1}{(x^{2} + x + 1)} \frac{d x}{\sqrt{x^{2} + x + 1}} = k \frac{x - 1}{\sqrt{x^{2} + x + 1}} + C$ then the value of $k$

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