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Why is it that integration of 1/(xlogx) is log(log x) and not log(log x) (the integration of log x)? See when we do the u substitution we get the answer but doesn't chain rule apply for integration? For eg- integration of (3(x)^2)/(1+(x)^2) We do u substitution and get in tegration of du/(1+u). My question is why is it log(1+x^3) +c and not log(1+x^3)*(x+x^4/4). So we don't integrate the inside term? Also what is the integration of log x.

amWhy
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geek101
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  • The integral of $\log x$ is $x\log x-x+C$. One can verify this by differentiating. One can also get to the result by integration by parts, $u=\log x$, $dv=dx$. – André Nicolas Aug 06 '14 at 14:51
  • $u$-substitution is the chain rule (in 'reverse'). – beep-boop Aug 06 '14 at 14:53
  • You are doing the chain rule with $u$-substitution, that's literally how the substitution works. But you cannot just say "I want to multiply by the integral of inner functions," just because you multiplied by derivatives in the derivative chain rule. Rules aren't pattern-matching--i.e. you cannot just change some words to get the rule for integrals vs derivatives--they work according to how the math makes them work. There are infinite ways which don't work for integrating something, only one that does. – Adam Hughes Aug 06 '14 at 14:54
  • And your first sentence doesn't make sense. – beep-boop Aug 06 '14 at 14:54

2 Answers2

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Let $u = \log x$. Then $\,du = \frac 1x \,dx$. We need to determine $\,du\,$ in order to take into account (reverse, so to speak) the use of the chain rule involved in differentiating the desired function.

Back to the integral: By substitution, we get

$$\int \frac 1{x\log x} \,dx = \int \frac 1{\log x}\cdot \frac 1x \,dx = \int \frac 1u \,du$$

This, in turn is equal to $\log|u| + C = \log|(\log x)| + C$.

Indeed, if we differentiate: $$\left(\log|\log(x)| + C\right)' = \frac 1{\log x} \cdot (\log x)' = \frac 1{\log x} \cdot \frac 1x = \frac 1{x\,\log x}$$

amWhy
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  • Please see my example. So we don't integrate the term inside log? Chain rule is not valid? Please read my whole question. – geek101 Aug 06 '14 at 14:51
  • When we differentiate the argument inside the first log, we get $(\log x)' = \frac 1x$. That's taken care of by the $x$ in the denominator of the integral. – amWhy Aug 06 '14 at 14:55
  • The u-substitution simplifies the integral. We are taking $(\log x)',dx$ into account: it represents $du$. Then we integrate, we integrate only the variable of substitution. Then, only after integrating, we replace what the variable of substitution represents, in this case, $u = \log x$. – amWhy Aug 06 '14 at 15:00
  • Please note that by using u-substitution, the aim is to take into account the chain-rule involved in the differentiation of a function (thinking of the integrand as the derivative of the function we need to find), and reversing (undoing) the chain-rule, so to speak, to recover a function prior to its differentiation. So it would be VERY incorrect to integrate what is already being integrated – amWhy Aug 06 '14 at 15:06
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We want to evaluate $$\int \frac{1} {x\ln(x)}dx.$$

Let $\color{green}{u=\ln(x)} \implies \frac{du}{dx}=\frac{1}{x} \iff \color{red}{du=\frac{1}{x}dx}$.

Then we have $$\int \frac{1}{\color{green}{\ln(x)}} \cdot \color{red}{\frac{1}{x}dx}=\int \frac{1}{\color{green}u}\color{red}{du}=\ln|u|+C=\ln|\ln(x)|+C$$

beep-boop
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  • Please see my example. So we don't integrate the term inside log? Chain rule is not valid? Please read my whole question. – geek101 Aug 06 '14 at 14:51