Integrate $\ln(t^2)/t$ with Substitution

[karush]

$\int\frac{\ln\left({t^2}\right)}{t}dt$

$\frac{1}{2}\int\ln\left({t}\right)\frac{1}{t}dt$

$u=\ln\left({t}\right)\ du=\frac{1}{t}dt$

$\frac{1}{2}\int u\ du$I continued but didn't get the answer

 

[jacobi1]

Hint: $\frac{\ln(t^2)}{t}= 2 \frac{\ln(t)}{t}$.
The rest of your evaluation is correct.

 

[Jameson]

$\ln(x^2) = 2\ln(x)$. You put the 2 in the denominator.

 

[karush]

So its.
$2\int\ u \ du$

$2 \left[\frac{u^2}{2}\right]+C$

 

[Jameson]

Yep, and the 2's would cancel.

 

[karush]

$(\ln\left({t}\right))^2+C$