Techniques of Integration

$\int \ln xdx=x\ln x-x+C$. 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.

Parts in the wrong direction. For $\int x{e}^{x}dx$, choosing $u=e^x$ and $dv=xdx$ gives

The identity is correct and the result is useless: the new integrand carries $x^2$ where the original had $x$. Repeating the choice raises the power again. Nothing has gone wrong except the direction of the trade, which is why the new integral must be inspected before continuing.

Partial fractions on an improper fraction. Attempting

produces no solution: the right side has degree $-1$ at best while the left tends to 1 as $x\to \mathrm{\infty }$. Clearing denominators gives $x^2=A(x+2)+B(x-1)$, a quadratic equal to a linear expression, which fails at the $x^2$ coefficient. Division first gives $1+\frac{-x+2}{(x-1)(x+2)}$, and the remainder is proper.

Partial fractions where the denominator does not factor. $\frac{1}{x^2+1}$ has an irreducible quadratic denominator over the reals, so no decomposition into linear pieces exists. Forcing one produces complex coefficients where a real answer was wanted; the integral is $\arctan x$, which no amount of splitting will reveal.

Substitution where no inner derivative is present. For $\int e^{x^2}dx$, setting $u=x^2$ needs $du=2xdx$, and there is no $x$ factor to supply it. The substitution cannot be completed. This integral has no elementary antiderivative at all, a theorem, not a failure of ingenuity, which is why the related $\int e^{-x^2}dx$ is handled numerically.

The contrast is $\int x{e}^{x^2}dx$, where the $x$ is present and the substitution works immediately, giving $\frac{1}{2}{e}^{x^2}$.

Concluding divergence from an unbounded integrand. ${\int }_0^1\frac{dx}{\sqrt{x}}$ has an integrand growing without bound as $x\to 0^{+}$, and it converges, to 2. The partial values $1.3675$, $1.9368$, $1.9980$ climb toward it. Unboundedness alone decides nothing: the rate decides it, and puts this case on the convergent side of the threshold.

Concluding convergence from a vanishing integrand. ${\int }_1^{\mathrm{\infty }}\frac{dx}{x}$ has an integrand tending to 0 and diverges. Tending to zero is necessary for convergence at infinity but nowhere near sufficient. The decay must beat $1/x$.

The two failures are mirror images, and together they are the reason the p-test is stated as a threshold rather than a rule of thumb.

Splitting that hides a divergence. For ${\int }_{-1}^1\frac{dx}{x^3}$, the two halves are $-\mathrm{\infty }$ and $+\mathrm{\infty }$, and pairing them symmetrically appears to give 0. That is a different quantity, the Cauchy principal value, and the integral itself diverges, because convergence requires each piece to converge independently.

An improper integral is evaluated by substituting $\mathrm{\infty }$ into the antiderivative, so ${\int }_1^{\mathrm{\infty }}\frac{dx}{x}=\ln \mathrm{\infty }-\ln 1$ is a calculation like any other.