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Evaluate the following integrals :\(\int\frac{log\,x}{x}\)dx |
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Answer» Assume, logx = t ⇒ d(logx) = dt ⇒ \(\frac{1}{x}\)dx = dt Substituting t and dt in above equation we get, ⇒ \(\int\)t.dt ⇒ \(\frac{t^2}{2}+c\) But, t = log(x) ⇒ \(\frac{log^2x}{2}\) + c . |
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