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The value of the integral $I=∫\frac{(logx)^3}{x}dx $ is : |
$\frac{1}{4}log(x^4)+C,$ where C is a constant $\frac{(logx)^4}{4}+C,$ where C is a constant $\frac{1}{2}log(x^2)+C,$ where C is a constant $\frac{1}{2}(logx)^4+C,$ where C is a constant |
$\frac{(logx)^4}{4}+C,$ where C is a constant |
The correct answer is Option (2) → $\frac{(logx)^4}{4}+C,$ where C is a constant $I=\int\frac{(\log x)^3}{x}dx$ Let $y=\log x$, $dy=\frac{dx}{x}$ $⇒I=\int y^3dy$ $⇒\frac{y^4}{4}+C=\frac{(\log x)^4}{4}+C$ |
