Integration II

Differentiation is a set of rules that always works. Integration is a set of techniques, and the skill is recognising which one an integral needs. That recognition is what this topic trains.

Standard integrals

$$\int e^x\,dx = e^x + C \qquad \int e^{kx}\,dx = \frac{e^{kx}}{k} + C$$

$$\int \frac1x\,dx = \ln|x| + C$$

This last one fills the gap left by the power rule at $n = -1$.

A pattern worth learning to spot:

$$\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$$

$$\int \frac{2x}{x^2+1}\,dx = \ln(x^2+1) + C$$

because $2x$ is exactly the derivative of $x^2+1$. (No modulus needed here — $x^2+1$ is always positive.)

Also worth recognising, from the inverse trigonometric derivatives:

$$\int\frac{dx}{1+x^2} = \tan^{-1}x + C \qquad \int\frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$$

Integration by substitution

The reverse of the chain rule. Choose $u$ to be the awkward inner function, then convert everything — including $dx$ — into terms of $u$.

Find $\displaystyle\int 2x(x^2+1)^5\,dx$.

Let $u = x^2+1$, so $\dfrac{du}{dx} = 2x$, giving $du = 2x\,dx$.

$$\int 2x(x^2+1)^5\,dx = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C$$

Find $\displaystyle\int x\sqrt{x^2+4}\,dx$.

Let $u = x^2+4$, so $du = 2x\,dx$ and $x\,dx = \frac12 du$:

$$\int x\sqrt{x^2+4}\,dx = \frac12\int u^{1/2}\,du = \frac12 \cdot \frac{2}{3}u^{3/2} + C = \frac13(x^2+4)^{3/2} + C$$

Integration by parts

The reverse of the product rule:

$$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$$

The whole art is choosing $u$. Pick the factor that gets simpler when differentiated.

Find $\displaystyle\int xe^x\,dx$.

Take $u = x$ (which differentiates to 1) and $\frac{dv}{dx} = e^x$ (so $v = e^x$):

$$\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C$$

Check by differentiating: $e^x(x-1) + e^x = xe^x$ ✓

Find $\displaystyle\int \ln x\,dx$.

There appears to be only one factor — so take $u = \ln x$ and $\frac{dv}{dx} = 1$, giving $v = x$:

$$\int \ln x\,dx = x\ln x - \int x \cdot \frac1x\,dx = x\ln x - x + C$$

Find $\displaystyle\int x\ln x\,dx$.

Here $u = \ln x$ again — a logarithm never becomes simpler when integrated, so it must be the part you differentiate. With $\frac{dv}{dx} = x$ and $v = \frac{x^2}{2}$:

$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac1x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$

Partial fractions

A fraction with a factorised denominator can be split into simpler pieces, each of which integrates to a logarithm.

Express $\dfrac{5x+1}{(x-1)(x+2)}$ in partial fractions.

$$\frac{5x+1}{(x-1)(x+2)} \equiv \frac{A}{x-1} + \frac{B}{x+2}$$

Multiplying up:

$$5x+1 \equiv A(x+2) + B(x-1)$$

Let $x = 1$: $\;6 = 3A$, so $A = 2$. Let $x = -2$: $\;-9 = -3B$, so $B = 3$.

$$\frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{3}{x+2}$$

Check by recombining: $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$ ✓

Now the integral is easy:

$$\int \frac{5x+1}{(x-1)(x+2)}\,dx = 2\ln|x-1| + 3\ln|x+2| + C$$

If the numerator's degree is greater than or equal to the denominator's, divide first, then apply partial fractions to the remainder.

The trapezium rule

When an integral cannot be found exactly, approximate it with strips:

$$\int_a^b y\,dx \approx \frac{h}{2}\Big[y_0 + 2(y_1 + y_2 + \cdots + y_{n-1}) + y_n\Big] \qquad h = \frac{b-a}{n}$$

The first and last ordinates are counted once; every ordinate in between is doubled.

Estimate $\displaystyle\int_0^1 x^2\,dx$ with 4 strips.

$h = 0.25$, and the ordinates are $0,\ 0.0625,\ 0.25,\ 0.5625,\ 1$:

$$\approx \frac{0.25}{2}\Big[0 + 2(0.0625 + 0.25 + 0.5625) + 1\Big] = 0.125\big[1.75 + 1\big] = 0.34375$$

The exact value is $\frac13 = 0.3333$, so the estimate is slightly high.