Rearranging integrals to enable $u$-Substitution

Using integration by subsitution for functions of the form $f(x) = g(u(x))$

In some cases, we can even apply integration by substitution when the $u'(x)$ term is missing. This trick can work if we can write $u'(x)$ as a function of $u(x)$ so that $u'(x) = h(u(x))$. If this is indeed the case, we can use the following algebra to make $f(x)$ fit a form which allows us to integrate by subsitution

$$\begin{aligned} f(x) &= g(u(x)) \\ f(x) &= g(u(x)) \cdot \frac{u'(x)}{u'(x)} \\ f(x) &= g(u(x)) \cdot \frac{u'(x)}{h(u(x))} \\ f(x) &= \frac{g(u(x))}{h(u(x))} \cdot u'(x) \\ f(x) &= g^\star(u(x)) \cdot u'(x) \quad \text{where } g^\star(u)=\frac{g(u)}{h(u)} \\ \end{aligned}$$

As an example, we will solve the integral $\int \frac{1}{e^x + 1} dx$

We could try using $u(x) = e^x$ to solve this integral. In this case, $u'(x) = e^x = u(x)$, so we could try writing

$$\begin{aligned} \int \frac{1}{e^x + 1} dx &= \int \frac{1}{e^x + 1} \cdot \frac{e^x}{e^x} dx \\ \int \frac{1}{e^x + 1} dx &= \int \frac{1}{(e^x + 1)e^x} \cdot e^x dx \\ \int \frac{1}{e^x + 1} dx &= \int \frac{1}{(u(x) + 1)u(x)} \cdot u'(x) dx \\ \int \frac{1}{e^x + 1} dx &= \int \frac{1}{(u + 1)u} du \quad \text{where } u = e^x \\ \end{aligned}$$

We can then solve this integral to obtain

$$\begin{aligned} \int \frac{1}{e^x + 1} dx &= \ln{|u|} - \ln{|u+1|} + C \\ \int \frac{1}{e^x + 1} dx &= \ln{|e^x|} - \ln{|e^x+1|} + C \\ \int \frac{1}{e^x + 1} dx &= x - \ln{(e^x+1)} + C \\ \end{aligned}$$

Common functions which can be useful for this integration trick are

$$\begin{aligned} u(x) &= e^{kx} &\quad u'(x) &= k \cdot u(x) \\ u(x) &= \tan(x) &\quad u'(x) &= u(x)^2 + 1 \\ u(x) &= \sqrt{x} &\quad u'(x) &= \frac2{u(x)} \\ u(x) &= \frac1{x} &\quad u'(x) &= -u(x)^2 \\ \end{aligned}$$

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