Integrate

Question

Integrate

dls · Answer

$$\Huge \int\limits_{}^{} \frac{\sqrt{16+(logx)^2}}{x}dx$$

dls · Answer

Attempt:
Let $16+(logx)^2=t$ 
$$dx=\frac{xdt}{2logx}$$
$$\LARGE\int\limits\limits_{}^{} \frac{\sqrt{t}}{ \cancel{x}}\frac{\cancel{x}dt}{2logx}=\frac{1}{2}\int\limits_{}^{} \frac{\sqrt{t}dt}{\sqrt{t-16}}=\int\limits_{}^{}\sqrt{\frac{t}{t-16}}dt$$

dls · Answer

just a try..:/

dls · Answer

umm @zepdrix

zepdrix · Answer

I think what you want to do is let $\large t=\ln x$
then, $\large dt=\dfrac{1}{x}$

Giving us,$$\large \int\limits\frac{\sqrt{16+(logx)^2}}{x}dx \qquad=\qquad \int\limits \sqrt{16+t^2}\quad dt$$From here we can do a trig sub i suppose.

zepdrix · Answer

$\large dt=\dfrac{1}{x}dx$ bah i forgot the dx when i wrote that :\

anonymous · Answer

assuming that your trail is correct , I complete by substituting $$t=16\sec ^{2}u$$ and it will develop to $$\int\limits\limits_{}^{}32\sec^2(u) du$$ which is 32tan(u)+c  still you have to find t in terms of u and blah blah blah.