Why Some Integrals Must Be Broken Into Pieces First

Many integrals resist both the standard table and a single substitution — fractions with quadratic denominators, products like $x{e}^x$, or square roots of quadratics. Each has its own tool: special formulae, partial fractions or integration by parts.

This part covers special integrals, partial fractions, integration by parts, and integrals of square roots of quadratic expressions.

**Six special integrals handle expressions involving $x^2+a^2$, $x^2-a^2$ or $a^2-x^2$, and a quadratic such as $x^2+bx+c$ is brought into one of these forms by completing the square.

The formulae:**

- $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$
- $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C$
- $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$
- $\int \frac{dx}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C$
- $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$
- $\int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C$

Worked example 1. $\int \frac{dx}{x^2+9}=\frac{1}{3}{\tan }^{-1}\frac{x}{3}+C$.

Worked example 2 — completing the square.

$$\int \frac{dx}{x^2+4x+13}=\int \frac{dx}{(x+2{)}^2+3^2}=\frac{1}{3}{\tan }^{-1}\frac{x+2}{3}+C$$

An everyday example. Rearranging a room so it matches a floor plan you already know is what completing the square does — it reshapes a quadratic to fit a known formula.

The substance. **Only $\sqrt{a^2-x^2}$ in the denominator leads to ${\sin }^{-1}$** — the forms with $\sqrt{x^2\pm a^2}$ give logarithms.

A proper rational function whose denominator factorises is split into simpler fractions with unknown constants, the constants are found by comparing numerators, and each simple fraction is integrated separately.

Standard forms for proper fractions:

- $\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$
- $\frac{px+q}{(x-a{)}^2}=\frac{A}{x-a}+\frac{B}{(x-a{)}^2}$
- $\frac{p{x}^2+qx+r}{(x-a)(x^2+bx+c)}=\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}$

Worked example. Evaluate $\int \frac{x+5}{(x+1)(x+2)}dx$.

$$\frac{x+5}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\Rightarrow x+5=A(x+2)+B(x+1)$$

Putting $x=-1$ gives $A=4$; putting $x=-2$ gives $3=-B$, so $B=-3$.

$$\int \left(\frac{4}{x+1}-\frac{3}{x+2}\right)dx=4\log |x+1|-3\log |x+2|+C$$

Improper fractions first. If the numerator's degree is not less than the denominator's, divide first to get a polynomial plus a proper fraction.

An everyday example. Splitting a group restaurant bill into each person's share turns one complicated amount into simple parts — just as partial fractions do.

The substance. A repeated factor needs one term for each power — $(x-a{)}^2$ needs both $\frac{A}{x-a}$ and $\frac{B}{(x-a{)}^2}$.

**Integration by parts integrates a product using $\int uvdx=u\int vdx-\int \left(u^{\prime }\int vdx\right)dx$, with $u$ chosen as the factor that simplifies on differentiation, and the special result $\int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C$ handles a common pattern at once.

Choosing the first function $u$. A helpful order is inverse trigonometric, logarithmic, algebraic, trigonometric, exponential.

Worked example 1.** $\int x{e}^{x}dx$, with $u=x$ and $v=e^x$:

$$\int x{e}^{x}dx=x{e}^x-\int 1\cdot e^{x}dx=x{e}^x-e^x+C$$

Worked example 2. Writing $\int \log xdx$ as $\int \log x\cdot 1dx$:

$$\int \log xdx=x\log x-\int \frac{1}{x}\cdot xdx=x\log x-x+C$$

**The $e^x[f(x)+f^{\prime }(x)]$ result.** For $\int e^x(\sin x+\cos x)dx$, take $f(x)=\sin x$, so $f^{\prime }(x)=\cos x$:

$$\int e^x(\sin x+\cos x)dx=e^x\sin x+C$$

An everyday example. Two people carrying a heavy trunk up stairs, one lifting while the other steadies and then swapping roles, mirrors how integration by parts shifts the work from one factor to the other.

The substance. **A poor choice of $u$ makes things harder** — taking $u=e^x$ in $\int x{e}^{x}dx$ leads to $\int \frac{x^2}{2}{e}^{x}dx$.

These three integrals come from integration by parts and are used as standard formulae, after rewriting any quadratic under the root in one of these forms by completing the square.

The formulae:

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

$$\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log \left|x+\sqrt{x^2-a^2}\right|+C$$

$$\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+C$$

Worked example 1. With $a=4$:

$$\int \sqrt{16-x^2}dx=\frac{x}{2}\sqrt{16-x^2}+8{\sin }^{-1}\frac{x}{4}+C$$

Worked example 2 — complete the square. Since $x^2+2x+5=(x+1{)}^2+2^2$,

$$\int \sqrt{x^2+2x+5}dx=\frac{x+1}{2}\sqrt{x^2+2x+5}+2\log \left|x+1+\sqrt{x^2+2x+5}\right|+C$$

An everyday example. The area of a semicircular flower bed comes from $\int \sqrt{a^2-x^2}dx$, because $y=\sqrt{a^2-x^2}$ traces the upper half of a circle.

The substance. **Only the $a^2-x^2$ form contains ${\sin }^{-1}$; the other two contain logarithms** — mixing them up is the usual error.