Question
$\int\frac{1}{\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3}}\text{dx}$
✓
Answer
$\int\frac{\text{dx}}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})}$
Rationalise the denominator
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}}-3)}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}\text{dx}$
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}{(2\text{x}+3)-(2\text{x}-3)}\text{dx}$
$=\frac{1}{6}\int(2\text{x}+3)^{\frac{1}{2}}\text{dx}-\frac{1}{6}\int(2\text{x}-3)^{\frac{1}{2}}\text{dx}$
$=\frac{1}{6}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]-\frac{1}{6}\Bigg[\frac{(2\text{x}-3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{1}{18}\big\{(2\text{x}+3)^{\frac{3}{2}}-(2\text{x}-3)^{\frac{3}{2}}\big\}+\text{c}$
Rationalise the denominator
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}}-3)}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}\text{dx}$
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}{(2\text{x}+3)-(2\text{x}-3)}\text{dx}$
$=\frac{1}{6}\int(2\text{x}+3)^{\frac{1}{2}}\text{dx}-\frac{1}{6}\int(2\text{x}-3)^{\frac{1}{2}}\text{dx}$
$=\frac{1}{6}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]-\frac{1}{6}\Bigg[\frac{(2\text{x}-3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{1}{18}\big\{(2\text{x}+3)^{\frac{3}{2}}-(2\text{x}-3)^{\frac{3}{2}}\big\}+\text{c}$
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