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Evaluate the indefinite integral. I am having trouble.


Help me. Please!

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

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

Directly: Using that

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

You have

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

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

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

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