Question

(i) Find the integral: $\int \frac{dx}{x^2-16}$

(ii) Find the integral: $\int \frac{dx}{\sqrt{2x-x^2}}$

Answer 1

(i) $\int \frac{dx}{x^2-16}$

$\int \frac{dx}{x^2-4^2}$

Using formula $\int \frac{dx}{x^2-a^2}$

$=\frac{1}{2a}\text{log}\left|\frac{x-a}{x+a}\right|+C$

$=\frac{1}{2\times 4}\text{log}\left|\frac{x-4}{x+4}\right|+C$

$=\frac{1}{8}\text{log}\left|\frac{x-4}{x+4}\right|+C$

Where $C$ is constant of integration.

(ii)$\int \frac{dx}{\sqrt{2x-x^2}}$

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

$=\int \frac{dx}{\sqrt{1-(x^2-2x+1)}}$

$=\int \frac{dx}{\sqrt{1-(x-1{)}^2}}=\int \frac{dx}{\sqrt{1^2-(x-1{)}^2}}$

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

$={\sin }^{-1}\left(\frac{(x-1)}{1}\right)+C$

$={\sin }^{-1}(x-1)+C$

Where $C$ is constant of integration.