Question

Solve : $I=\int \frac{dx}{\sqrt{3x+4}-\sqrt{3x+1}}$

Answer 1

$I=\int \frac{dx}{\sqrt{3x+4}-\sqrt{3x+1}}$

$t=\frac{1}{\sqrt{3x+4}-\sqrt{3x+1}}\frac{1}{t}=\sqrt{3x+4}-\sqrt{3x+1}$

$t=\frac{\sqrt{3x+4}+\sqrt{3x+1}}{(3x+4-3x-1)}$

$t=\frac{1}{3}(\sqrt{3x+4}+\sqrt{3x+1})$

$\Rightarrow 3t+\frac{1}{t}=2\sqrt{3x+4}$

$\Rightarrow \left[\frac{1}{2}\right[3t+\frac{1}{4}]{]}^2=3x+4$

$\frac{1}{4}\times 2(3t+\frac{1}{t})\times (3-\frac{1}{t^2})dt=3dx$

$dx=\frac{1}{6}\left(\frac{3t^2+1}{t}\right)\left(\frac{3t^2-1}{t^2}\right)dt$

$dx=\frac{1}{6}\left(\frac{9t^2-1}{t^3}\right)dt$

$I=\int t\times \frac{1}{6}\frac{(9t^4-1)}{t^3}dt$

$=\int \frac{1}{6}\frac{(9t^4-1)}{t^2}dt$

$=\frac{1}{6}\int 9t^{2}dt=\frac{1}{6}\int \frac{1}{t^2}dt$

$=\frac{3}{2}\frac{t^3}{3}-\frac{1}{6}\frac{t^{2+1}}{(-2+1)}+c$

$=I=\frac{t^3}{2}+\frac{1}{6t}+c$

$=I=\frac{2}{2\sqrt{3x+4}-\sqrt{3x+1}3}+\frac{1}{6}(\sqrt{3x+4}-\sqrt{3x+1})+6$