Preparing Expressions for Integration | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Rewrite roots and reciprocals as powers

The power rule acts on single powers. A square root has power $1/2$ ; a reciprocal changes the sign of the power. Multiplying powers adds their indices, while dividing subtracts them.

\begin{gathered}\sqrt{x}=x^{1/2},\quad \frac1{x^2}=x^{-2}\\\frac1{\sqrt{x}}=x^{-1/2},\quad x\sqrt{x}=x^{3/2}\end{gathered}

Worked example

Integrate 6x² − 4/x³, for x ≠ 0.

\begin{gathered}\int(6x^2-4x^{-3})\,dx\\=\frac63x^3-\frac4{-2}x^{-2}+c\\=2x^3+\frac2{x^2}+c\end{gathered}

The second coefficient is positive because $-4\div(-2)=2$ . Differentiating $2x^{-2}$ gives $-4x^{-3}$ .

For a fractional divisor, divide by multiplying by its reciprocal: $5\div\tfrac12=10$ . Keep the original domain: a denominator cannot be zero, and the expression inside a real square root must be non-negative.

Expand products and divide each term

You can integrate a sum term by term, but you cannot integrate two factors separately and multiply the answers. Expand a product first when that gives a short sum of powers.

\begin{gathered}(x+1)(x-2)=x^2-x-2\\\int(x+1)(x-2)\,dx\\=\frac{x^3}{3}-\frac{x^2}{2}-2x+c\end{gathered}

When a single power divides a numerator, divide every term by it. For $x>0$ :

\begin{gathered}\frac{x^2+2x-3}{\sqrt{x}}\\=x^{3/2}+2x^{1/2}-3x^{-1/2}\end{gathered}

\begin{aligned}&\int\frac{x^2+2x-3}{\sqrt{x}}\,dx\\&=\frac25x^{5/2}+\frac43x^{3/2}\\&\quad-6\sqrt{x}+c\end{aligned}

Your turnPractice

Find $\int(12x^3+6/x^2-5/\sqrt{x})\,dx$ for $x>0$ , and check the signs by differentiating.

Show worked answer

Rewrite the integrand as $12x^3+6x^{-2}-5x^{-1/2}$ .

Answer

3x^4-\frac6x-10\sqrt{x}+c

The derivative of $-6x^{-1}$ is positive; the derivative of $-10x^{1/2}$ is negative, as required.

Your turnChoose the first step

Find $\int (x+2)^2/\sqrt{x}\,dx$ , with $x>0$ .

Show worked answer

Expand the numerator, then divide each term by $x^{1/2}$ .

\begin{gathered}\frac{(x+2)^2}{\sqrt{x}}\\=x^{3/2}+4x^{1/2}+4x^{-1/2}\end{gathered}

Answer

\begin{aligned}&\frac25x^{5/2}+\frac83x^{3/2}\\&\quad+8\sqrt{x}+c\end{aligned}

Rewrite roots and reciprocals as powers

The power rule acts on single powers. A square root has power $1/2$ ; a reciprocal changes the sign of the power. Multiplying powers adds their indices, while dividing subtracts them.

\begin{gathered}\sqrt{x}=x^{1/2},\quad \frac1{x^2}=x^{-2}\\\frac1{\sqrt{x}}=x^{-1/2},\quad x\sqrt{x}=x^{3/2}\end{gathered}

Worked example

Integrate 6x² − 4/x³, for x ≠ 0.

\begin{gathered}\int(6x^2-4x^{-3})\,dx\\=\frac63x^3-\frac4{-2}x^{-2}+c\\=2x^3+\frac2{x^2}+c\end{gathered}

The second coefficient is positive because $-4\div(-2)=2$ . Differentiating $2x^{-2}$ gives $-4x^{-3}$ .

Expand products and divide each term

You can integrate a sum term by term, but you cannot integrate two factors separately and multiply the answers. Expand a product first when that gives a short sum of powers.

\begin{gathered}(x+1)(x-2)=x^2-x-2\\\int(x+1)(x-2)\,dx\\=\frac{x^3}{3}-\frac{x^2}{2}-2x+c\end{gathered}

When a single power divides a numerator, divide every term by it. For $x>0$ :

\begin{gathered}\frac{x^2+2x-3}{\sqrt{x}}\\=x^{3/2}+2x^{1/2}-3x^{-1/2}\end{gathered}

\begin{aligned}&\int\frac{x^2+2x-3}{\sqrt{x}}\,dx\\&=\frac25x^{5/2}+\frac43x^{3/2}\\&\quad-6\sqrt{x}+c\end{aligned}

Your turnPractice

Find $\int(12x^3+6/x^2-5/\sqrt{x})\,dx$ for $x>0$ , and check the signs by differentiating.

Show worked answer

Rewrite the integrand as $12x^3+6x^{-2}-5x^{-1/2}$ .

Answer

3x^4-\frac6x-10\sqrt{x}+c

The derivative of $-6x^{-1}$ is positive; the derivative of $-10x^{1/2}$ is negative, as required.

Your turnChoose the first step

Find $\int (x+2)^2/\sqrt{x}\,dx$ , with $x>0$ .

Show worked answer

Expand the numerator, then divide each term by $x^{1/2}$ .

\begin{gathered}\frac{(x+2)^2}{\sqrt{x}}\\=x^{3/2}+4x^{1/2}+4x^{-1/2}\end{gathered}

Answer

\begin{aligned}&\frac25x^{5/2}+\frac83x^{3/2}\\&\quad+8\sqrt{x}+c\end{aligned}