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