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Calculus BC · Unit 6 · 6.14
AP Calculus · Integration and Accumulation of Change
Selecting Techniques for Antidifferentiation
The hard part is no longer executing a technique — it is choosing one. Classify by structure.
key idea$\text{Read the integrand's \emph{shape}, not its difficulty}$
How to choose
By this point you own several antidifferentiation techniques, and the hard part is no longer executing one — it is picking one. Choosing is a matter of recognising structure: a composition, a product, a top-heavy fraction, a factorable denominator. Spend ten seconds classifying before you commit.
| If the integrand looks like… | The structural clue | Technique |
|---|
| a basic form — $x^{n}$, $e^{x}$, $\sin x$, $\dfrac1x$ | you already know its antiderivative | recognise it |
| $f\!\left(g(x)\right)\cdot g'(x)$ | an inner function and its derivative both present | $u$-substitution |
| a sum, or a fraction with a monomial denominator | the terms are independent | split it up and integrate term by term |
| top degree $\ge$ bottom degree | an improper rational function | long division first |
| $\dfrac{1}{\text{irreducible quadratic}}$ | negative discriminant, no matching numerator | complete the square → $\arctan$ |
| $\dfrac{N(x)}{(x-a)(x-b)}$ | the denominator factors into distinct linears | partial fractions |
| a product of two unrelated functions | no inner-derivative match anywhere | integration by parts |
Read the rows top-down: the cheapest recognitions come first. Always test for substitution before reaching for anything heavier — it dissolves more integrals than every other technique combined.
The decision tree
Work the branches in order. Each one is a cheap yes/no test, and the first ‘yes’ decides the problem. If nothing fires, the integrand almost certainly needs algebraic rewriting before any technique will apply.
FIRSTIs it already a form you have memorised?Power rule, exponential, the six trig antiderivatives, $\dfrac1x\to\ln\lvert x\rvert$, or an inverse-trig form. Write it down and stop. No technique needed.
THENIs there an inner function whose derivative appears (up to a constant)?Use $u$-substitution. A missing constant factor is fine — adjust with a multiplier. A missing variable factor is not, and means substitution alone will fail.
THENCan it be split or simplified by algebra?Distribute a product, split a sum over a single denominator, cancel a common factor, or apply a trig identity such as $\sin^{2}x=\tfrac{1-\cos2x}{2}$. Then re-enter the tree at the top.
RATIONALIs it a rational function?Compare degrees. Top-heavy → long division. Denominator factors into distinct linears → partial fractions. Irreducible quadratic → complete the square for an $\arctan$ or $\arcsin$ form.
PRODUCTIs it a product of two structurally unrelated functions?Such as $x e^{x}$ or $x\ln x$ — substitution cannot touch these. Use integration by parts.
STUCKNothing fires?Re-read the integrand for a hidden rewrite: multiply by a conjugate, add and subtract a term, or convert every trig function to sine and cosine. On the AP exam a definite integral that resists all of this is meant to be done numerically on the calculator.
Test firstsubstitution — it solves the most integrals
Test secondalgebra — splitting and identities beat techniques
Neverreach for a heavy tool before the cheap tests
Worked examples
EX 1
Routing to substitution
inner derivative presentChoose a technique for $\displaystyle\int x\sqrt{x^{2}+1}\,dx$ and evaluate it.
1
Basic form?
No — the composition $\sqrt{x^{2}+1}$ is not on the memorised list.
2
Inner derivative present?
Inner function $x^{2}+1$, derivative $2x$. The integrand has $x$ — off by the constant $2$, which is fine.
3
Substitute
$u=x^{2}+1$, $du=2x\,dx$, so $x\,dx=\tfrac12du$.
4
Integrate
$\tfrac12\displaystyle\int u^{1/2}du=\tfrac13u^{3/2}$.
5
Undo it
$\dfrac13\left(x^{2}+1\right)^{3/2}+C$. The tree stopped at branch two.
EX 2
Routing to algebra
split, do not substituteChoose a technique for $\displaystyle\int\dfrac{x^{2}+3}{x}\,dx$.
1
Test for substitution
Inner function $x$? Its derivative is $1$, which gives nothing. There is no useful inner function here.
2
Test for algebra
The denominator is a single monomial, so the fraction splits: $\dfrac{x^{2}}{x}+\dfrac{3}{x}=x+\dfrac3x$.
3
Re-enter the tree
Both pieces are now basic forms.
4
Integrate
$\dfrac{x^{2}}{2}+3\ln\lvert x\rvert+C$.
5
The lesson
Splitting is legal because the denominator is one term. Splitting across a sum in the denominator is not — $\dfrac{1}{a+b}\neq\dfrac1a+\dfrac1b$.
EX 3
Routing to completing the square
rational, irreducibleChoose a technique for $\displaystyle\int\dfrac{dx}{x^{2}+6x+13}$.
1
Test for substitution
Derivative of the denominator is $2x+6$; the numerator is $1$. No match, so substitution alone fails.
2
Enter the rational branch
Numerator degree $0$, denominator degree $2$ — proper, so no division.
3
Check whether it factors
Discriminant $36-52=-16\lt 0$. Irreducible, so partial fractions are unavailable.
4
Complete the square
$x^{2}+6x+13=(x+3)^{2}+4$, giving $\displaystyle\int\dfrac{du}{u^{2}+4}$ with $u=x+3$.
5
Finish
$\dfrac12\arctan\dfrac{x+3}{2}+C$. Three tests were eliminated before the right one appeared.
EX 4
A trap: the near-substitution
why the tree order mattersExplain why $\displaystyle\int\sqrt{x^{2}+1}\,dx$ cannot be routed the way Example 1 was.
1
Compare with Example 1
Same inner function $x^{2}+1$, but the outside factor $x$ is gone.
2
Attempt the substitution
$u=x^{2}+1$ needs $du=2x\,dx$. There is no $x$ to supply it.
3
Reject the illegal fix
Writing $dx=\dfrac{du}{2x}$ leaves an $x$ in the integral. You may never carry the original variable into a $u$-integral.
4
Note the true difficulty
This integral requires trigonometric substitution — outside the AP course entirely.
5
The takeaway
‘Looks like’ a substitution is not enough. The derivative must actually be present, up to a constant.
EX 5
Routing a definite integral
check the interval tooChoose a technique for $\displaystyle\int_{0}^{1}\dfrac{2x}{x^{2}+1}\,dx$.
1
Test for substitution
Derivative of $x^{2}+1$ is $2x$, which is exactly the numerator. Branch two fires immediately.
2
Substitute with limits
$u=x^{2}+1$: when $x=0$, $u=1$; when $x=1$, $u=2$. The integral becomes $\displaystyle\int_{1}^{2}\dfrac{du}{u}$.
3
Check continuity on the interval
$x^{2}+1$ never vanishes, so the integrand is continuous on $[0,1]$ and the FTC applies.
4
Evaluate
$\left[\ln u\right]_{1}^{2}=\ln2$.
5
Why convert the limits
Changing them with the variable removes the need to substitute back — and forgetting to do either is the classic definite-integral error.
Watch out
A constant may be adjusted; a variable may not
$\int x\sqrt{x^{2}+1}\,dx$ works, $\int\sqrt{x^{2}+1}\,dx$ does not. The difference is one factor of $x$ — and it decides everything.
Simplify before you classify
Distributing a product or splitting a fraction often turns a hard-looking integrand into two basic forms. Try algebra before any technique.
$\dfrac{1}{a+b}\neq\dfrac1a+\dfrac1b$
Fractions split over a monomial denominator only. A sum in the denominator requires factoring and partial fractions, not splitting.
Difficulty is not the signal — shape is
An intimidating integrand with a visible inner derivative is easy; a short one without it may be impossible. Classify by structure, never by how it looks.