I apologise in advance if this does not meet post guidelines.

I am having difficulty with U-Substitution. I cannot seem to find an answer anywhere.

Okay, so (if I'm not mistaken) u-substitution can be used whenever the integral is of the form: $$\color{lime}{\int(f(g(x)) * (g'(x))\,dx} $$ where $\color{lime}{u=g(x)}$ .



BUT I've noticed that everyone seems to solve the following integral using u-substitution. $$\color{red}{\int\frac{(\arctan(x))}{1+x^2}\,dx}$$ I don't understand how it works on this integral, because it appears to be of the form: $$\color{red}{\int(g(x))*(g'(x))\,dx}$$ where $\color{red}{u=g(x)=\arctan(x)}$,

instead of $\color{lime}{\int(f(g(x)) * (g'(x))dx} $



I am just so confused. I recognise that, in the $\color{red}{\arctan(x)}$ example, $\color{lime}{f(x)}$ (of the green formula) could merely be $f(x)=x$, but u-substitution does not seem to work on other examples where $f(x)=x$. For example in: $$\color{blue}{\int{e^{3x}}\,dx}$$ I know this could obviously be solved simply by recognising that this is the integral of an exponential function, and hence is equal to $\color{blue}{\frac{1}{3}e^{3x}+c}$

BUT, if it does hold true that u-substitution can still be performed even when f(x) of the green function is just $f(x)=x$, then surely this COULD be solved using u-substitution once it is rearranged to give: $$\color{blue}{\frac{1}{3e^{3x}}\int{e^{3x}}*{3e^{3x}}dx}$$ because the integral is now of the form $\color{blue}{\int(f(g(x)))*(g'(x))dx}$ where $\color{blue}{f(x)=x}$

BUT when I try to simplify this, I end up with $$\color{blue}{\frac{e^{3x}}{6}}$$ instead? So clearly u-substitution does NOT always work when $\color{lime}{f(x)}$ of the green formula is $f(x)=x$??

Ugh, I'm so confused. Any help would be GREATLY appreciated. I am rather pressed for time, as I have an assignment that I need to submit by tonight. May I please also request that explanations be explained in such a way a numpty like me can understand.

Thank you in advance.

  • $\begingroup$ It just means $f(x)=x$. $\endgroup$ – Simply Beautiful Art Jun 19 '19 at 9:56
  • $\begingroup$ Also too many unnecesary colours, hard to read. $\endgroup$ – Simply Beautiful Art Jun 19 '19 at 9:57
  • $\begingroup$ For the $\color{blue}{\text{blue}}$ part, in $$\color{blue}{\int e^{3x}\cdot\frac{3e^{3x}}{3e^{3x}}dx}$$ the $\color{blue}{\dfrac{1}{3e^{3x}}}$ is not a constant and cannot be pulled out of integration. $\endgroup$ – peterwhy Jun 19 '19 at 9:59
  • $\begingroup$ Thank you for your reply. Sorry about the colours -- I thought it might help to organise things. I know in the arctan example that f(x)=x. But why doesn't u-substitution work for the exponential example? $\endgroup$ – user642965 Jun 19 '19 at 10:01
  • 1
    $\begingroup$ Yes? u-substitution always works as long as you do it right and follow the necessary conditions...? Even when $f(x)=x$. $\endgroup$ – Simply Beautiful Art Jun 19 '19 at 10:06

For the $\color{blue}{\text{blue}}$ part, where

$$\color{blue}{\int e^{3x}dx = \int e^{3x}\cdot\frac{3e^{3x}}{3e^{3x}}dx}$$

the $\color{blue}{\dfrac{1}{3e^{3x}}}$ is not a constant and cannot be pulled out of integration. So

$$\color{blue}{\int e^{3x}\cdot\frac{3e^{3x}}{3e^{3x}}dx \ne \frac{1}{3e^{3x}}\int e^{3x}\cdot 3e^{3x} dx}$$

But what if we let $g(x) = e^{1.5x}$?

$$\begin{align*} \int e^{3x}dx &= \int e^{1.5x}e^{1.5x}dx\\ &= \frac23 \int e^{1.5x}\cdot 1.5e^{1.5x} dx\\ &= \frac23 \int e^{1.5x}d\left(e^{1.5x}\right)\\ &= \frac23 \frac{\left(e^{1.5x}\right)^2}{2}+C\\ &= \frac{e^{3x}}{3}+C \end{align*}$$

  • $\begingroup$ Oh of course!!!! No wonder it wasn't working haha. Thank you very much, peterwhy :) $\endgroup$ – user642965 Jun 19 '19 at 10:07

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