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Evaluate:

Question:

Evaluate: $\int \frac{1}{\sqrt{\mathrm{x}+\mathrm{a}}+\sqrt{\mathrm{x}+\mathrm{b}}} \mathrm{dx}$

Solution:

Let $I=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$

$=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$

Now, Multiply with conjugate, we get

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

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

$=\int \frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b} d x$

$=\frac{1}{a-b}\left[\frac{2}{3}(x+a)^{\frac{2}{2}}-\frac{2}{3}(x+b)^{\frac{2}{2}}\right]$

Hence, $\mathrm{I}=\frac{2}{3(\mathrm{a}-\mathrm{b})}\left[(\mathrm{x}+\mathrm{a})^{\frac{3}{2}}-(\mathrm{x}+\mathrm{b})^{\frac{3}{2}}\right]+\mathrm{C}$

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