# Regarding u-substitution

1) $\displaystyle \int_{1}^{4} \frac{(\ln x)^3}{2x}dx$
2) $\displaystyle \int_{}^{} \frac{\ln(\ln x)}{x \ln x}dx$
3) $\displaystyle \int \frac{e^{\sqrt{r}}}{\sqrt{r}}dr$
4) $\displaystyle \int \frac{\ln(x)}{x \sqrt{/\ln^2x+1}}dx$

Regarding these questions, I am a bit confused bit u-substitution. For 1), I tried substituting $ln(x)^3$ with u but made it even messier. If anyone could show me how to best approach these questions, opposed to a full answer, and how to look for u-substitution, that would be great.

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Hints to get you started:

1) $\displaystyle \int_{1}^{4} \frac{(\ln x)^3}{2x}dx$

$\quad \displaystyle u = \ln x \implies du = \frac{dx}{x} = \frac{1}{x}\,dx$

2) $\displaystyle \int_{}^{} \frac{\ln(\ln x)}{x \ln x}dx$

$\quad \displaystyle u = \ln(\ln x)\implies du = \frac{dx}{x \ln x} = \frac 1{x \ln x}\,dx$

3) $\displaystyle \int_{}^{} \dfrac{e^{\sqrt r}}{\sqrt r} dr$

$\quad \displaystyle u = \sqrt r \implies du = -\frac 12\cdot \frac{dr}{\sqrt r}\iff -2\,du = \frac{dr}{\sqrt r} = \frac{1}{\sqrt r}\,dr$

4) $\displaystyle \int_{}^{} \dfrac{(\ln x)dx}{x \sqrt{\ln^2x+1}}$

$\quad \displaystyle u = \ln^2 x + 1 \implies du = 2\frac{\ln x\,dx}{x} = 2\cdot \frac{ln x}{x}\,dx$

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Nice hints - +1 – Amzoti Jun 2 '13 at 20:51

Hint:

$$\frac{dx}{x} = d(\ln{x})$$

$$\frac{dx}{x \ln{x}} = d(\ln{\ln{x}})$$

$$\int f'[g(x)] \, d[g(x)] = f[g(x)] + C$$

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"Jhbjk Mbm", you seem to be missing the point that it's all about the chain rule, and that that tells you what kind of substitution to look for.

You have: $$\int (\ln x)^2\,\frac{dx}{2x}$$

Here's a hint: $$\int (\ln x)^2 \frac12 \Big(\frac1x\,dx\Big)$$

What that's hinting at is that the thing in the big parentheses is $du$. And that should tell you what $u$ should be.

What you need to look for in $u$-substitutions is something (in this case $1/x$) that when multiplied by $dx$, is the differential of some other expression that appears in the integral.

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All these integrals are doable with the following easy-to-check fact:

If $\,f\,$ is a derivable function and $\,g\,$ is an (indefinitely) integrable function, then

$$\int f'(x)g(f(x))\,dx =G(f(x))\;,\;\;\text{where \,G\, is a primitive function of}\;\;g$$

For example, in case (2), we have

$$f(x)=\log \log x\;,\;g(x)= x\implies$$

$$\implies \int f'(x)g(f(x))\,dx=\int\log(\log x)\cdot\frac1{x\log x}dx=\frac12\log^2\log x$$

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