# Evaluate:

Question:

Evaluate: $\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} \mathrm{dx}$

Solution:

Let I $=\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x$

$I=\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x$

Now, Multiply with the conjugate, we get

$=\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} \times \frac{(\sqrt{2 x+3}-\sqrt{2 x-3})}{\sqrt{2 x+3}-\sqrt{2 x-3}} d x$

$=\int \frac{(\sqrt{2 x+3}-\sqrt{2 x-3})}{(\sqrt{2 x+3})^{2}-(\sqrt{2 x-3})^{2}} d x$

$=\int \frac{(\sqrt{2 x+3}-\sqrt{2 x-3})}{2 x+3-2 x+3} d x$

$=\int \frac{\sqrt{2 x+3}}{6} d x-\int \frac{\sqrt{2 x-3}}{6} d x$

$=\frac{1}{6} \int(2 \mathrm{x}+3)^{\frac{1}{2}} \mathrm{~d} \mathrm{x}-\frac{1}{6} \int(2 \mathrm{x}-3)^{\frac{1}{2}} \mathrm{dx}$

$=\frac{1}{6}\left(\frac{2 \times+3}{2}\right)^{\frac{1}{2}+1}-\frac{1}{6}\left[\frac{2 \times-3}{2}\right]^{\frac{1}{2}+1}$

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

Hence, $I=\frac{1}{18}(2 x+3)^{\frac{2}{2}}-\frac{1}{18}(2 x-3)^{\frac{2}{2}}+C$