Question
$\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\text{dx}.$ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\times\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}\text{dx}$
$=\int\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\text{x}+3-\text{x}-2}\text{dx}$
$=\int\Big[(\text{x}+3)^{\frac{1}{2}}+(\text{x}+2)^{\frac{1}{2}}\Big]\text{dx}$
$=\frac{(\text{x}+3)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(\text{x}+2)^{\frac{3}{2}}}{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\times(\text{x}+3)^{\frac{3}{2}}+\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
$\text{I}=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
$\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\times\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}\text{dx}$
$=\int\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\text{x}+3-\text{x}-2}\text{dx}$
$=\int\Big[(\text{x}+3)^{\frac{1}{2}}+(\text{x}+2)^{\frac{1}{2}}\Big]\text{dx}$
$=\frac{(\text{x}+3)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(\text{x}+2)^{\frac{3}{2}}}{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\times(\text{x}+3)^{\frac{3}{2}}+\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
$\text{I}=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
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