Integral of log with an exponential inside. ln(x^13)

Question

I was wondering as to how to integrate logs with exponentials inside of them. Could you help me understand how this can be solved? I know the best way is to do it by parts, but I am not sure which variable to substitute. $$\int_3^4 ln(x^{13}) dx$$

Answer 1

Hint:

$$\ln(x^{13})=13\ln(x)$$

if $x>0$.

Answer 2

It's a simple property of $\ln$ that if you have $\ln(a^x)$, where $x>0$ (which it is in this case), then that is the same as $x \ln(a)$. So you can rewrite your integral as:

$$\int^{4}_{3}(13\ln(x))dx$$ $$13\int^{4}_{3}\ln(x)dx$$

Then just find the indefinite integral of $\ln(x)$ which is $x\ln \left(x\right)-x$.

Evaluate at boundaries.

$$[4\ln(4)-4]-[3\ln(3)-3] = 4\ln(4)-4-3\ln(3)+3=4\ln(4)-3\ln(3)-1$$

But remember we gotta multiply our answer by $13$.

So, we get:

$$52\ln(4) - 39\ln(3) - 13$$

Answer 3

Hint:

Rewrite the integral as \begin{align} 13\int^4_3\ln x\ dx \end{align} then use integration by parts where $u = \ln x$.