Question

Find $\int \frac{dx}{x^2-a^2}$ and hence evaluate $\int \frac{dx}{3x^2+13x-10}$.

Answer 1

To find $\int \frac{dx}{x^2-a^2}$

Let us first expand the term given under integral sign i.e. $\frac{1}{x^2-a^2}$

$\frac{1}{x^2-a^2}=\frac{1}{(x-a)(x+a)}$

Let $\frac{1}{(x-a)(x+a)}=\frac{A}{(x-a)}+\frac{B}{(x+a)}$

$⟹A(x+a)+B(x-a)=1$

$⟹A+B=0$ and $A-B=\frac{1}{a}$

$⟹B=-\frac{1}{2a}$ and $A=\frac{1}{2a}$

$⟹\int \frac{dx}{x^2-a^2}=\int [\frac{1}{2a(x-a)}-\frac{1}{2a(x+a)}]dx$

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

$=\frac{1}{2a}\left[\log \right(x-a)-\log (x+a)$

$=\frac{1}{2a}\log \left(\frac{x-a}{x+a}\right)$

Consider $\frac{1}{3x^2+13x-10}=\frac{1}{(3x-2)(x+5)}$

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

$⟹A(x+5)+B(3x-2)=1$

$⟹A+3B=0$ and $5A-2B=1$

$⟹A=\frac{3}{17}$ and $B=-\frac{1}{17}$

$\therefore \int \frac{1}{3x^2+13x-10}=\int [\frac{3}{17(3x-2)}-\frac{1}{17(x+5)}]dx$

$=\frac{1}{17}\left[3\log \right(3x-2)-\log (x+5\left)\right]$

$=\frac{1}{17}\left[\log \right(3x-2{)}^3-\log (x+5)]$

$=\frac{1}{17}\log \left(\frac{(3x-2{)}^3}{(x+5)}\right)$

Hence, $\int \frac{1}{3x^2+13x-10}=\frac{1}{17}\log \left(\frac{(3x-2{)}^3}{(x+5)}\right)$