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Evaluate ` int x log x dx`. |
Answer» `int underset (II)(x)underset(I)(log)x dx` `=logx {int x dx}-int {(d)/(dx)(logx)intxdx}dx` `=(logx)(x^(2))/(2)int(1)/(x)(x^(2))/(2)dx` `=(x^(2))/(2)log x-(1)/(2)int x dx` `=(x^(2))/(2)logx-(1)/(2)((x^(2))/(2))+C=(x^(2))/(2)logx-(1)/(4)x^(2)+C` |
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