Evaluate the indefinite integral. I am having trouble.


Help me. Please!



Let $u=\ln 7x$, then $du=\frac 1xdx$.

  • $\begingroup$ $du \ne \frac{1}{x} dx$ You forgot the inner function, "$7x$" which has to be differentiated as well. $\endgroup$ – Zhoe Dec 3 '13 at 21:13
  • $\begingroup$ Differentiating $\ln 7x$ yields $\frac 1{7x}(7x)'=\frac 1{7x}7=\frac 1x$... $\endgroup$ – abiessu Dec 3 '13 at 21:19
  • $\begingroup$ Oh whoops, sorry. Forgot the $7$ in the denominator..+1! $\endgroup$ – Zhoe Dec 3 '13 at 21:20
  • $\begingroup$ This identity, namely that $\frac {d(\ln ax)}{dx}=\frac 1x$, is also proven by the fact that $\ln ax=\ln a+\ln x$. $\endgroup$ – abiessu Dec 3 '13 at 21:38
  • $\begingroup$ Oh, I see, makes sense..forgive me, I haven't done differentiation in a while, so I am a lot rusty.. $\endgroup$ – Zhoe Dec 3 '13 at 21:41

Directly: Using that

$$\int \frac{f'(x)}{f(x)}dx=\log f(x)+C$$

You have

$$(\log 7x)'=\frac1x\implies\int\frac{dx}{x\log 7x}=\int\frac{(\log 7x)'}{\log 7x}dx=\log\log 7x+C$$


$$\int\frac{dx}{x\ln7x}=\int\frac{dx}{x}\cdot\frac{1}{\ln7x}=|\ln7x=t\Rightarrow \frac{t}{7}dt=\frac{dx}{x}|$$ $$=\frac{1}{7}\int\frac{dt}{t}=\frac{1}{7}\ln |t|=\frac{1}{7}\ln |\ln7x|+C$$

  • 1
    $\begingroup$ I see a missing factor of $7$ somewhere... $\endgroup$ – abiessu Dec 3 '13 at 21:22
  • $\begingroup$ There should be no factor of 1/7 in front of the log term. $\endgroup$ – omegadot Jan 17 '16 at 13:37

substitute $\ \ln\left(7x\right)=u$, then $dx=xdu$

$$\int_{ }^{ }\frac{1}{u}du$$


undo substitution $\ \ln\left(7x\right)=u$: $$=\ln\left(\left|\ln\left(7x\right)\right|\right)+C.$$


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