Techniques of Integration

The integral unit reduced ${\int }_a^{b}f$ to finding a function that differentiates to $f$. For powers, sums and the odd substitution that is enough. For most integrands it is not.

$\int x{e}^{x}dx$ has no substitution: neither factor is the other's derivative. $\int \frac{dx}{(x-1)(x+2)}$ is a single opaque fraction, though the answer is a sum of two logarithms. And ${\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}$ asks for an accumulation over an interval that never ends, which is not a Riemann sum at all. There are infinitely many rectangles to add.

Each obstruction has its own technique. Parts inverts the product rule and shifts the differentiation from one factor to the other. Partial fractions splits a rational integrand into pieces the power and logarithm rules already reach. And an improper integral is defined as a limit of ordinary ones, which is what makes the question of convergence meaningful.

That last is not a formality. ${\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}$ converges to exactly 1, while ${\int }_1^{\mathrm{\infty }}\frac{dx}{x}$ diverges. Both integrands tend to zero, and only one does so fast enough. This unit also settles the case the integral unit flagged and left open: ${\int }_{-1}^1{x}^{-2}dx$, where applying the fundamental theorem across the singularity produced a finite negative number for an integral that is positive and unbounded.

The canonical definition states the three techniques and the p-test. What follows is why each carries the conditions it does.

Why the choice of $u$ decides whether parts helps. The formula $\int udv=uv-\int vdu$ is an identity whichever way the factors are assigned, so it is always valid and usually useless. For $\int x{e}^{x}dx$ with $u=x$, the new integral is $\int e^{x}dx$, simpler, because differentiating $x$ gave 1. Assigning $u=e^x$ instead gives $\int \frac{x^2}{2}{e}^{x}dx$, with a higher power than before. The identity held; the trade went the wrong way.

The useful ordering is a rough hierarchy for $u$: logarithms first, then inverse trigonometric functions, then powers of $x$, then exponentials and trigonometric functions. What it encodes is which functions simplify on differentiation and which are unchanged by it.

Why a logarithm is integrated by parts at all. $\int \ln xdx$ has one factor, so there is nothing to trade, until $dv=dx$ is chosen, making $v=x$ and $u=\ln x$. The method then gives $x\ln x-\int x\cdot \frac{1}{x}dx=x\ln x-x$. Parts applies whenever a function is easier to differentiate than to integrate, even alone.

Why partial fractions needs a proper function. The decomposition writes a fraction as a sum of terms each of degree $-1$ or lower. If the original has numerator degree at least the denominator's, no such sum can reproduce it. The equation for the coefficients is inconsistent. Polynomial division extracts the polynomial part first, leaving a proper remainder the method can handle.

Why the coefficients are determined. Clearing denominators turns the decomposition into an equality of polynomials, and two polynomials agree for all $x$ exactly when their coefficients match. That gives as many equations as unknowns. Substituting the roots of the denominator is a shortcut: each root kills all but one term, isolating one coefficient directly.

Why an improper integral is defined rather than evaluated. A Riemann sum needs a bounded interval cut into finitely many pieces of finite height. With an unbounded interval there is no finite partition, and with an unbounded integrand no finite rectangle height. So ${\int }_1^{\mathrm{\infty }}$ is not an integral whose value is being computed. It is notation for a limit, and the limit is where the meaning lives.

This is why $\ln \mathrm{\infty }-\ln 1$ is not a calculation: $\mathrm{\infty }$ is not a number the antiderivative can be evaluated at. The correct statement is $\underset{t\to \mathrm{\infty }}{lim}(\ln t-\ln 1)$, which does not exist.

Why an interior singularity forces a split. The limit definition approaches the bad point from one side. A singularity inside the interval has two sides, each needing its own limit, and the integral converges only if both do. Integrating straight through applies the fundamental theorem where its hypothesis fails.

Why the p-test has two different thresholds. At infinity the integrand must decay fast enough, so large $p$ helps: ${\int }_1^{\mathrm{\infty }}{x}^{-p}$ converges for . Near zero the integrand blows up, so large $p$ hurts: ${\int }_0^1{x}^{-p}$ converges for . The case $p=1$ fails at both ends, because $\ln x$ is unbounded in both directions, which is why $1/x$ is the standard boundary against which other integrands are compared.

Integration by parts. Start from the product rule for differentiable $u$ and $v$:

Integrate both sides over $[a,b]$. The left side is the derivative of $uv$, so by the fundamental theorem its integral is ${\left[uv\right]}_a^b$:

Rearranging isolates one integral:

which in differential notation is $\int udv=uv-\int vdu$. The method is the product rule with one term moved across, and the minus sign comes from that move rather than from any convention.

Immediate check. For ${\int }_0^{1}x{e}^{x}dx$ take $u=x$ and $dv=e^{x}dx$, so $du=dx$ and $v=e^x$:

Differentiating the antiderivative $x{e}^x-e^x$ returns $e^x+x{e}^x-e^x=x{e}^x$, and a midpoint sum with 400,000 subintervals.

The p-test at infinity. For $p\ne 1$ the power rule gives an antiderivative directly:

The behaviour as $t\to \mathrm{\infty }$ is decided by the sign of the exponent $1-p$:

The excluded case $p=1$ has antiderivative $\ln x$, and $\ln t\to \mathrm{\infty }$, so it diverges too. Checking the values: $\ln 10=2.303$, $\ln 100=4.605$, $\ln {10}^4=9.210$, $\ln {10}^6=13.816$, growing without bound, though slowly.

Verification of the convergent side. For $p=2$ the formula gives $\frac{1}{2-1}=1$. Directly, ${\int }_1^t{x}^{-2}dx=1-\frac{1}{t}$, which reads $0.9$ at $t=10$, $0.99$ at $t=100$, $0.9999$ at $t={10}^4$ and $0.999999$ at $t={10}^6$.

The p-test at zero. The same antiderivative, different endpoint:

Now $s\to 0^{+}$, so $s^{1-p}\to 0$ when and blows up when . The threshold reverses: convergence for , divergence for .

Verification. For $p=\frac{1}{2}$: ${\int }_s^1{x}^{-1/2}dx=2-2\sqrt{s}$, which reads $1.3675$ at $s=0.1$, $1.9368$ at $s=0.001$ and $1.9980$ at $s={10}^{-6}$, approaching 2. The region is unbounded in height and has finite area. The integrand's blow-up is slow enough that the accumulation stays finite.

Step 0 — check whether the integral is improper. Before any technique, ask two questions: is either limit infinite, and does the integrand blow up anywhere on the closed interval, endpoints included? A yes to either means the integral is defined as a limit, and Step 5 applies. Missing this is how a divergent integral acquires a finite answer.

Step 1 — try substitution. The cheapest method. Its signature is an inner function whose derivative appears as a factor, up to a constant. If it works, nothing else is needed.

Step 2 — for a product of unlike factors, use parts. The signature is a product whose factors come from different families: a power times an exponential, a power times a logarithm, a lone logarithm. Choose $u$ by which factor simplifies when differentiated:

Apply $\int udv=uv-\int vdu$ and inspect the new integral. If it is worse than the original, swap the choice; if both choices are worse, parts is the wrong method.

Step 3 — for a rational function, use partial fractions.

1. Check it is proper: numerator degree strictly below denominator degree. If not, divide first.
2. Factor the denominator completely.
3. Write one term per factor, following the table in the definition.
4. Clear denominators and solve, substituting each root isolates one coefficient at a time.
5. Integrate term by term; linear denominators give logarithms.

Step 4 — verify by differentiating. Every antiderivative found by any method can be checked by differentiating it back to the integrand. This catches sign errors from parts and arithmetic errors in the coefficients, and costs a few seconds.

Step 5 — for an improper integral, replace the bad endpoint with a variable and take a limit.

• Unbounded interval: ${\int }_a^{\mathrm{\infty }}f=\underset{t\to \mathrm{\infty }}{lim}{\int }_a^{t}f$.
• Unbounded at an endpoint $c$: approach it, $\underset{t\to c}{lim}{\int }_a^{t}f$.
• Unbounded at an interior point: split there first, then treat each piece separately. Both must converge, or the whole diverges.

Evaluate the proper integral in $t$ by the techniques above, then take the limit. Report a finite value as convergence, and otherwise divergence with the reason.

Step 6 — use the p-test as a shortcut and a sanity check. For an integrand behaving like $x^{-p}$: at infinity it converges when ; near zero when . Comparing a complicated integrand with the nearest power often settles convergence before any antiderivative is attempted.

1. Parts with a power and an exponential: ${\displaystyle {\int }_0^{1}x{e}^{x}dx}$.

No substitution is available, neither factor is the other's derivative. Take $u=x$ (simplifies when differentiated) and $dv=e^{x}dx$ (easy to integrate), so $du=dx$ and $v=e^x$:

Check: differentiating $x{e}^x-e^x$ gives $e^x+x{e}^x-e^x=x{e}^x$, and a midpoint sum returns.

The other choice. Taking $u=e^x$ and $dv=xdx$ gives $\frac{x^2}{2}{e}^x-\int \frac{x^2}{2}{e}^{x}dx$. The new integrand carries $x^2$ where the original had $x$. The identity still holds; the trade went the wrong way.

2. Parts on a lone logarithm: ${\displaystyle {\int }_1^{2}x\ln xdx}$.

Take $u=\ln x$ and $dv=xdx$, so $du=\frac{dx}{x}$ and $v=\frac{x^2}{2}$:

So the value is $2\ln 2-(1-\frac{1}{4})=2\ln 2-\frac{3}{4}\approx \mathbf{0.6362943611}$.

Check: a midpoint sum.

The logarithm was chosen as $u$ because differentiating it produces $1/x$, which cancels against the $x^2$ from $v$. That cancellation is why the method works here.

3. Partial fractions: ${\displaystyle {\int }_3^5\frac{dx}{(x-1)(x+2)}}$.

The integrand is proper, numerator degree 0, denominator degree 2, so decompose directly:

Substituting the roots isolates each coefficient: at $x=1$, $1=3A$ so $A=\frac{1}{3}$; at $x=-2$, $1=-3B$ so $B=-\frac{1}{3}$.

Verify the identity at $x=3$: the left side is $\frac{1}{2\cdot 5}=0.1$, and the right is $\frac{1/3}{2}-\frac{1/3}{5}=0.1$.

Integrating term by term:

Check: a midpoint sum returns.

4. A convergent improper integral: ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}}$.

The interval is unbounded, so this is notation for a limit:

Check: the partial values are $0.9$ at $t=10$, $0.99$ at $t=100$, $0.9999$ at $t={10}^4$, $0.999999$ at $t={10}^6$. The p-test agrees: , value $\frac{1}{p-1}=1$.

5. A divergent one: ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{x}}$.

which does not exist, $\ln 10=2.30$, $\ln 100=4.61$, $\ln {10}^4=9.21$, $\ln {10}^6=13.82$, growing without bound. The integral diverges.

Compare case 4: both integrands tend to 0, and only $x^{-2}$ does so fast enough for the accumulated area to stay finite. $p=1$ is exactly the threshold.

$\int \ln xdx=x\ln x-x+C$. A single factor, integrated by parts with $dv=dx$. Memorise it with the reason: the logarithm is easy to differentiate and has no elementary antiderivative in sight, so it must be the $u$.

$\int x{e}^{x}dx=x{e}^x-e^x+C$. The template for a power times an exponential. Each application of parts lowers the power by one, so $\int x^2{e}^{x}dx$ takes two passes and $\int x^3{e}^{x}dx$ three.

$\int x\ln xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$. Over $[1,2]$ this gives $2\ln 2-\frac{3}{4}\approx 0.63629$. The $1/x$ from differentiating the logarithm cancels against a power of $x$, which is what makes the remaining integral elementary.

A repeated linear factor: ${\displaystyle \frac{3x+5}{(x-1{)}^2}}={\displaystyle \frac{3}{x-1}}+{\displaystyle \frac{8}{(x-1{)}^2}}$. A repeated factor contributes one term per power, not a single term. Checking at $x=2$: the left side is $\frac{11}{1}=11$, and the right is $3+8=11$; at $x=4$ both give $1.888\bar{8}$. Integrating gives $3\ln |x-1|-\frac{8}{x-1}+C$. A logarithm from the first term and a power from the second.

${\int }_0^{\mathrm{\infty }}{e}^{-x}dx=1$. The standard convergent improper integral. The partial values climb $0.632$, $0.993$, $0.99995$, $0.999999998$ at $t=1,5,10,20$: exponential decay is fast enough that almost all the area sits near the origin. This is the normalisation behind the exponential distribution.

${\int }_1^{\mathrm{\infty }}{\displaystyle \frac{dx}{x^2+1}}={\displaystyle \frac{\pi }{4}}$. The antiderivative is $\arctan x$, and the limit is $\frac{\pi }{2}-\frac{\pi }{4}\approx 0.78540$. The partial values run $0.6857$, $0.7754$, $0.7853$ at $t=10,100,{10}^4$. Convergence could have been predicted without any antiderivative: the integrand is smaller than $x^{-2}$, whose integral converges.

${\int }_0^1{\displaystyle \frac{dx}{\sqrt{x}}}=2$. Improper at the left endpoint, where the integrand is unbounded. The partial values are $1.3675$, $1.9368$, $1.9980$ as $s\to 0^{+}$. Finite area under an unbounded curve, which is the case worth carrying: unboundedness alone never decides convergence.

Three are parts, two are the decomposition, three are improper, and the improper ones split two to one between convergent and divergent for the same reason each time, the rate at which the integrand approaches its bad point. Reading that rate is what the p-test formalises.