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\(\int\frac{1+x^2+ln\,x}{x+x^2ln\,x}dx\) |
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Answer» \(\int\frac{1+x^2+ln\,x}{x+x^2ln\,x}dx\) Let x+ x2ln x = t ⇒ (1 + \(\frac{x^2}x\) + 2x ln x) dx = dt ⇒ (1 + x (1 + 2 ln x))dx = dt \(\therefore\) \(\int\frac{[1+x(1+2ln\,x)]dx}{x+x^2ln\, x}\) = \(\int\frac{dt}t=ln\,t + c\) = ln|x + x2ln x| + c (\(\because\) t = x + x2 ln x) |
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