When to use Integral Substitution?

Question

$e^x$$(1+e^x)^{1\over{2}}$

why can't use integration by part,

What is meant by in the form of f(g(x))g'(x)? Can you give a few example?

Thank you

Answer 1

Integration by substitution says that $$ \int f(g(x))g'(x) \; dx = \int f(u)\;du $$ where $u = g(x)$. This is a tool that will let you compute the integral of all functions that look exactly like $f(g(x))g'(x)$. When you first see this, you might be thinking that this seems like a very specialised rule that only seem good for very special cases. But it turns out that this is very useful.

In your example, you have $$ \int (1 + e^x)^{1/2}e^x\; dx. $$ If you want to know whether or not you can use substitution you just have to determine if it is possible to write $(1 + e^x)^{1/2}e^x$ as $f(g(x))g'(x)$ for some functions $f$ and $g$. Mabe by trail and error you find that indeed if $f(x) = x^{1/2}$ and $g(x) = 1+e^x$, then $f(g(x))g'(x) = (1 + e^x)^{1/2}e^x$. And so with $u = g(x) = 1+e^x$, you get $$ \int (1 + e^x)^{1/2}e^x\; dx = \int u^{1/2} \; du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(1+e^x)^{3/2} + C. $$

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You ask why you you can't use integration by parts. Integration by parts says that $$ \int f(x)g'(x)\; dx = f(x)g(x) - \int f'(x)g(x)\; dx. $$ So you would have to find $f$ and $g$ such that $f(x)g'(x) = e^x(1+e^e)^{1/2}$. You can of course choose $f(x) = (1 + e^x)^{1/2}$ and $g(x) = e^x$, so then you get $$ (1 + e^x)^{1/2}e^x - \int \frac{1}{2}(1+e^x)^{1/2}e^xe^x\;dx. $$ This is perfectly fine, but it doesn't do you any good. You wanted to find one complicated integral and now you just have another complicated integral. So you can use integration by parts, but, again, it doesn't help you much. It would be a good exercise to try other choices of $f$ and $g$ to convince yourself that this indeed doesn't help much.

Answer 2

in your example i would set $t=\sqrt{1+e^{x}}$ then we get $t^2=1+e^{x}$ and we have $2tdt=e^xdx$ thus your integral is $2\int t^2dt$

Answer 3

$$\int (e^x+1)^{1/2}e^x\ dx=\int (1+\color{green}{e^x})^{1/2}\ d\color{green}{e^x}=\int(1+\color{green}{u})^{1/2}\ d\color{green}{u}=\frac23(1+u)^{3/2}=\frac23(1+e^x)^{3/2}.$$

Equivalenty,

$$\int (e^x+1)^{1/2}e^x\ dx=\int (\color{green}{1+e^x})^{1/2}\ d(\color{green}{1+e^x})=\int \color{green}{u}^{1/2}\ d\color{green}{u}=\frac23u^{3/2}=\frac23(1+e^x)^{3/2}.$$

Note that $dg(x)$ is the same as $g'(x)\ dx$.