Evaluating a definite integral

Question

Evaluate the definite integral. $$\int_3^7 \ln(x^{43})\,dx$$

Answer 1

$$\begin{align} \int_3^7 dx \: \log{x^{43}} &= 43 \int_3^7 dx \: \log{x}\\ &= 43 [(x \log{x} - x)]_3^7 \\ &= 43 (7 \log{7} - 7 - (3 \log{3} - 3)) \end{align}$$

Answer 2

HINT: $\ln(x^a)=a\ln(x)$ This should do the trick if you really wanted $\ln(x^{43})$

Answer 3

hint: $\int_3^7 \ln x^{43} \, dx=\int_3^7 43 \ln x\,dx$ and compute following integral on [3,7] $$\int \ln x =x\ln x-x$$

Answer 4

Recall the properties of logarithms, in particular:

$$\;\ln(a^b) = b \ln a\;\implies \;\int_3^7 \; \ln({x^{43}})\,dx \;= \; \int_3^7 43\ln{x}\,dx\; = \;43\int_3^7 \ln x\,dx$$ Then use the fact that $\;\;\displaystyle43\int_3^7\,\ln x\,dx\;=\; 43[x\ln x-x]\Big|_3^7 \;$,

and evaluate, given your bounds of integration!

Answer 5

This is sort of a joke answer (but correct).

If we do not know any "log laws," let $u=\log(x^{43})$, $dv=dx$. Then $du=\dfrac{43x^{42}}{x^{43}}\,dx=\dfrac{43}{x}\,dx$, and we can take $v=x$. Thus our indefinite integral is $$x\log(x^{43})-\int \frac{43}{x} x\,dx.$$