# u-substitution problem help

• Nov 27th 2008, 01:33 PM
Matt164
u-substitution problem help
I need help with a u-substitution problem. It is has already been worked out for me but I don't understand how to get the final answer.
Here is the problem:

∫ln(x)/x

here is the worked out answer:

u=ln(x) ---> ∫udu=(u^2)/2 + C
=(ln(x))^2)/2 + C

I don't understand why ln(x) changes to u^2 when you substitute u into the integral, and I also don't understand where the 2 on the bottom comes from. I thought the substitution should be u/x if you substitute u for ln(x) into the integral. Maybe a step was skipped in the solution and that's why I don't understand? Anyway, if anyone could explain to me why it becomes u^2/2 + C instead of u/x, I would greatly appreciate it. Thanks in advance to anyone who can help.
• Nov 27th 2008, 01:42 PM
Chris L T521
Quote:

Originally Posted by Matt164
I need help with a u-substitution problem. It is has already been worked out for me but I don't understand how to get the final answer.
Here is the problem:

∫ln(x)/x

here is the worked out answer:

u=ln(x) ---> ∫udu=(u^2)/2 + C
=(ln(x))^2)/2 + C

I don't understand why ln(x) changes to u^2 when you substitute u into the integral, and I also don't understand where the 2 on the bottom comes from. I thought the substitution should be u/x if you substitute u for ln(x) into the integral. Maybe a step was skipped in the solution and that's why I don't understand? Anyway, if anyone could explain to me why it becomes u^2/2 + C instead of u/x, I would greatly appreciate it. Thanks in advance to anyone who can help.

When you make a substitution, you are changing the integral into something you can easily integrate. In integrating $\int\frac{\ln(x)}{x}\,dx$, you make the substitution ${\color{blue}u=\ln(x)}$. Differentiating, we get $\frac{\,du}{\,dx}=\frac{1}{x}\implies {\color{red}\,du=\frac{\,dx}{x}}$

Now let's rexamine the integral:

$\int\frac{{\color{blue}\ln(x)}}{{\color{red}x}}{\c olor{red}\,dx}$

Making the proper substitutions, we end up with $\int u\,du$

Integrating, we end up with $\tfrac{1}{2}u^2+C$ (we need a constant since this is indefinite integration [no limits])

But we were after $\int\frac{\ln(x)}{x}\,dx$, so now backsubstitute $u=\ln(x)$ back into your antiderivate to now get $\color{red}\boxed{\tfrac{1}{2}\left(\ln(x)\right)^ 2+C}$

Does this process make a little more sense now?