Indefinite integral of ( lnx / x + 4x(lnx)² dx
Can someone help me with this?? IDK where to start even. I was thinking about letting u = ln(x), du = 1/x then I ended up with.............(1/4) integral (u/u²)du
$\displaystyle \int \frac{\ln{x}}{x + 4x(\ln{x})^2} \, dx$
$\displaystyle \int \frac{\ln{x}}{1 + 4(\ln{x})^2} \cdot \frac{1}{x} \, dx$
$\displaystyle u = \ln{x}$ ... $\displaystyle du = \frac{1}{x} \, dx$
$\displaystyle \int \frac{u}{1 + 4u^2} \, du$
can you finish?
Oh I see my mistake. I added the 4 and the 1 ... and took (1/5) out of the intagral.I understand how you got to that now (both of you) Now I do integration by parts? It would be nice if 1 of you can actually do it out step by step, because I often mess up on this.
Or...I just saw an easier way, The integral can become [ (u/1) + (u/4u²) ] du and then this is very simple to integrate...is this correct?