Q.Find the integral $\dfrac{1}{x^2 - a^2}$ with respect to $x$ and hence evaluate $\int \dfrac{1}{x^2 - 16}\,dx$.

Standard integral: $\int\tfrac{dx}{x^2-a^2}=\tfrac{1}{2a}\log\left|\tfrac{x-a}{x+a}\right|+k$.

Using partial fractions, $\dfrac{1}{x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)$.

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Partial fractions give $\frac{1}{2}\log\left|\frac{x}{x+2}\right|+c$.

Write $\dfrac{1}{x(x+2)} = \dfrac{1}{2}\left(\dfrac{1}{x} - \dfrac{1}{x+2}\right)$.

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$\int \frac{dx}{x^2-1} = \frac{1}{2}\log\left|\frac{x-1}{x+1}\right| + k$.

The standard result is $\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + k$.

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Partial fractions give $\dfrac{x}{(x-1)(x-2)} = \dfrac{-1}{x-1}+\dfrac{2}{x-2}$, integrating to $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$.

Let $\dfrac{x}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$

$x = A(x-2)+B(x-1)$

At $x=1$: $1 = -A \Rightarrow A=-1$

At $x=2$: $2 = B \Rightarrow B=2$

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Clear the denominator and compare coefficients (or plug in $x=3$) to isolate $B$.

Write $\dfrac{1}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}$.

Multiplying both sides by $x(x-3)$:

$1 = A(x-3) + Bx$

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This is a standard result obtained by partial fractions: $\dfrac{1}{x^2-a^2} = \dfrac{1}{2a}\left(\dfrac{1}{x-a} - \dfrac{1}{x+a}\right)$.

$\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\int\left(\frac{1}{x-a} - \frac{1}{x+a}\right)dx = \frac{1}{2a}\left[\log|x-a| - \log|x+a|\right] + C$

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Partial fractions give $a=1,\ b=-1$, hence $a+b=0$.

Let $t=\sin x$. Then $\dfrac{1}{(1+t)(2+t)}=\dfrac{a}{1+t}+\dfrac{b}{2+t}$, so $1=a(2+t)+b(1+t)$.

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