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