More Substitution

"More substitution" simply means the same renaming trick as last lesson, used on integrals that need a bit of extra help first. All we're really doing is spotting familiar shapes. The formulas above are two shapes worth memorizing: $\frac{g'(x)}{g(x)}$ (a fraction whose top is the derivative of its bottom) always integrates to $\ln|g(x)|$, and $\frac{1}{a^2 + x^2}$ (a constant plus $x^2$ in the bottom) always gives an arctan. Once you can see those, most of these integrals are short.

Why does derivative-over-function give a log? Think about the chain rule on $\ln|g(x)|$: the derivative is $\frac{1}{g(x)}\cdot g'(x)$. That's exactly $\frac{g'(x)}{g(x)}$. So it's the substitution rule with $u = g(x)$ and $\int\frac{du}{u} = \ln|u| + C$.

Solve for $x$. In $\int x\sqrt{x + 1}\,dx$, let $u = x + 1$; the leftover $x$ becomes $u - 1$.

Logs in disguise. If the numerator is the derivative of the denominator (up to a constant), the answer is a log: $\int\tan x\,dx = \int\frac{\sin x}{\cos x}dx = -\ln|\cos x| + C$. Here the top, $\sin x$, is minus the derivative of the bottom, which is where the minus sign comes from.

Arctans in disguise. $\int\frac{dx}{1 + 4x^2}$: with $u = 2x$ it becomes $\frac12\int\frac{du}{1 + u^2} = \frac12\arctan(2x) + C$.

Split fractions. $\displaystyle\int\frac{3x - 1}{x^2 + 4}dx = \int\frac{3x}{x^2 + 4}dx - \int\frac{dx}{x^2 + 4}$: a log plus an arctan.

Integrals of the form $\int\frac{dx}{a^2 + x^2}$ give the arctan shapes in electric fields of charged rods, and log integrals appear whenever something's rate is proportional to $\frac1x$, like the work in expanding a gas.