$\displaystyle \int \dfrac{(\ln x)^{3}}{2x}dx$
hint?
Why would you want to "modify" a number? And why did you not answer my questions?
Your original integral was $\displaystyle \int \frac{(ln(x))^3}{2x}dx$ and skeeter suggested that you make the substitution u= ln(x) so that du= (1/x)dx.
So $\displaystyle \int\frac{(ln(x))^3}{2x}dx= \frac{1}{2}\int (ln(x))^3[(1/x)dx]= \frac{1}{2}\int u^3 du$
You should know that no such substitution will "move" dx into the denominator. There simply is no such thing as "$\displaystyle \int \frac{f(u)}{du}$"!
$\displaystyle \int \dfrac{(\ln x)^{3}}{2x}dx$
$\displaystyle u = \ln x$
$\displaystyle \dfrac{1}{2}du = \dfrac{1}{x} dx$ - Some wrong turn is made here. ???
$\displaystyle \frac{1}{2} \int u^3 \, du$
$\displaystyle \dfrac{1}{2} \dfrac{u^{4}}{4} + C$
$\displaystyle \dfrac{u^{4}}{8} + C$
$\displaystyle \dfrac{\ln(x)^{4}}{8} + C$
The book says the answer is $\displaystyle 2(\ln x) + C$, though ???
NO! All that is happening is 1/2 is being taken out of the integral as a factor. It is NOT a part of the substitution!
Well your book is clearly wrong. What do you get if you differentiate 2 ln(x) ? You should get 2/x, which is clearly not the same as your integrand...$\displaystyle \frac{1}{2} \int u^3 \, du$
$\displaystyle \dfrac{1}{2} \dfrac{u^{4}}{4} + C$
$\displaystyle \dfrac{u^{4}}{8} + C$
$\displaystyle \dfrac{\ln(x)^{4}}{8} + C$
The book says the answer is $\displaystyle 2(\ln x) + C$, though ???