1.

\( \int \frac{x-1}{\sqrt{x^{2}+1}} d x \)

Answer»
(√1+x²) - tan-¹x +c

\(\int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int\frac{x}{\sqrt{x^2 + 1}}dx - \int\frac{1}{\sqrt{x^2 + 1}}dx\)

Let \(x^2 + 1 = t^2\)

Then 

\(2x \,dx = 2t \,dt\)

⇒ \(x\,dx = t\,dt\)

\(\therefore \int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int \frac{t\,dt}t - log\left|x + \sqrt{x^2 +1}\right|+ C\)

\(= t - log \left|x + \sqrt{x^2 + 1}\right| + C\)

\(= \sqrt{x^2 + 1} - log \left|x + \sqrt{x^2 + 1}\right| + C\)



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