Integration by Substitution — the Workhorse Method

If $u = g(x)$, then $du = g'(x)\,dx$, and $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$. The art is choosing $g(x)$ so that its derivative is already present (up to a constant) in the integrand.

If the integrand is a fraction whose numerator is the derivative of its denominator, then $\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$. Often you must engineer the numerator: split it into 'a constant times $f'(x)$' plus a leftover, then integrate each piece.

If $u = f(x)$ then $\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C$ for $n \neq -1$ (and the $n = -1$ case is the log pattern). The shape $\displaystyle\int \dfrac{f'(x)}{[f(x)]^2}\,dx = -\dfrac{1}{f(x)} + C$ is the $n = -2$ instance.

For integrals containing $\sqrt{ax+b}$, substitute $t = \sqrt{ax+b}$ (so $x$ and $dx$ are expressed via $t$), or substitute $u = ax+b$. For $\sqrt{x}$ shapes, $t = \sqrt{x}$ gives $x = t^2,\ dx = 2t\,dt$. The radical disappears and the integrand becomes rational/polynomial in $t$.

Two faces of the same idea:

• Take-out-the-power: for $\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}$-type integrals, pull the dominant power out of the root — $(x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}$ — then substitute $u = 1 + x^{-4}$.
• Reciprocal $t = x \pm \dfrac{k}{x}$: for a denominator quadratic in $x^2$ with a matching numerator, divide top and bottom by $x^2$. Then $\dfrac{d}{dx}\!\left(x\mp\dfrac{k}{x}\right) = 1\pm\dfrac{k}{x^2}$ is exactly the new numerator, and the integral becomes a standard $t^2+1$ arctan (or a clean power).

For integrands built from $e^x$: substitute $t = e^x$, giving $dt = e^x\,dx$; a stray $e^x$ in the numerator becomes $dt$ and the rest becomes rational in $t$. For exponential towers $\int a^{a^x}\,a^x\,dx$, substitute $u = a^x$ (then $du = a^x \log a\,dx$).