Question

$\int \frac{1}{\sqrt{3x+5}-\sqrt{3x+2}}dx=$

Answer 1

We have,

$\int \frac{1}{\sqrt{3x+5}-\sqrt{3x+2}}dx$

On rationalize and we get,

$\int \frac{1}{\sqrt{3x+5}-\sqrt{3x+2}}\times \frac{\sqrt{3x+5}+\sqrt{3x+2}}{\sqrt{3x+5}+\sqrt{3x+2}}dx$

$=\int \frac{\sqrt{3x+5}+\sqrt{3x+2}}{{\left(\sqrt{3x+5}\right)}^2-{\left(\sqrt{3x+2}\right)}^2}dx$

$=\int \frac{\sqrt{3x+5}+\sqrt{3x+2}}{3x+5-3x-2}dx$

$=\frac{1}{3}\int \sqrt{3x+5}dx+\frac{1}{3}\int \sqrt{3x+2}dx$

$=\frac{1}{3}{\int (3x+5)}^{\frac{1}{2}}dx+\frac{1}{3}{\int (3x+2)}^{\frac{1}{2}}dx\therefore \int {(ax+b)}^n=\frac{{(ax+b)}^{n+1}}{a(n+1)}$

$=\frac{1}{3}\frac{{(3x+5)}^{\frac{1}{2}+1}}{3(\frac{1}{2}+1)}+\frac{1}{3}\frac{{(3x+2)}^{\frac{1}{2}+1}}{3(\frac{1}{2}+1)}$

$=\frac{1}{9}\frac{{(3x+5)}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{9}\frac{{(3x+2)}^{\frac{3}{2}}}{\frac{3}{2}}$

$=\frac{2}{9}[{(3x+5)}^{\frac{3}{2}}+{(3x+2)}^{\frac{3}{2}}]$

Hence, this is the answer.