Integration Techniques: Substitution, Parts & Partial Fractions

Three workhorse techniques turn almost any integral into ones you already know: u-substitution (reverse chain rule), integration by parts (reverse product rule), and partial fractions (break a rational function into simple pieces). Together they handle area, arc-length, and variable-limit problems mechanically.

1. Substitution (reverse chain rule)

Pick an inner function $u=g(x)$ so that $du=g'(x)\,dx$ already appears (up to a constant): $\int f\big(g(x)\big)\,g'(x)\,dx=\int f(u)\,du.$ For a definite integral either back-substitute at the end, or convert the limits directly: $x:a\to b$ becomes $u:g(a)\to g(b)$.

Example. $\displaystyle\int_0^1 x\,e^{x^2}\,dx$: let $u=x^2$, $du=2x\,dx$, giving $\tfrac12\int_0^1 e^u\,du=\tfrac12(e-1)$.

2. Integration by parts (reverse product rule)

$\int u\,dv=uv-\int v\,du.$ Choose $u$ by LIATE (Logarithmic, Inverse-trig, Algebraic, Trig, Exponential — whichever appears earlier is the one you differentiate). A key trick: to integrate a function whose derivative is simpler than itself, take $dv=dx$ so $v=x$. This is exactly how you integrate a variable-limit integral $F(x)=\int_a^x f(t)\,dt$: by the FTC $F'(x)=f(x)$, so $\int F(x)\,dx=x\,F(x)-\int x\,f(x)\,dx,$ and over $[a,\,b]$ the boundary term collapses because $F(a)=0$.

3. Partial fractions (rational functions)

Any proper rational $\dfrac{P(x)}{Q(x)}$ (with $\deg P<\deg Q$; otherwise long-divide first) splits over the factors of $Q$:

• a linear factor $(x-a)$ contributes $\dfrac{A}{x-a}$, which integrates to $A\ln|x-a|$;
• a repeated factor $(x-a)^k$ contributes $\dfrac{A_1}{x-a}+\cdots+\dfrac{A_k}{(x-a)^k}$;
• an irreducible quadratic $(x^2+px+q)$ contributes $\dfrac{Bx+C}{x^2+px+q}$, which integrates to a $\ln$ plus an $\arctan$.

The quadratic case rests on the standard integral $\displaystyle\int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac{x}{a}+C$.

Which technique to reach for

Mini-example (area between curves)

The region enclosed by $y=\dfrac{8}{x^2+4}$ and the parabola $y=\dfrac{x^2}{4}$ needs two different techniques at once: $\displaystyle\int\frac{8}{x^2+4}\,dx=4\arctan\frac{x}{2}$ (rational/arctan) and $\displaystyle\int\frac{x^2}{4}\,dx=\frac{x^3}{12}$ (power rule). Compute $\int_a^b(\text{top}-\text{bottom})\,dx$, and split the interval at each crossing where the upper and lower curves swap; exploiting even symmetry (integrate on $[0,b]$ and double) saves work.

Exam trick: For $\int_a^b F(x)\,dx$ with $F(x)=\int_a^x f$, never chase a closed form for $F$ — integrate by parts with $dv=dx$; the FTC converts $F'$ into $f$ and the $F(a)=0$ boundary term vanishes. And whenever a denominator looks like $x^2+a^2$, reach for $\tfrac1a\arctan\tfrac{x}{a}$, not a logarithm.