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Integrate the function
$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x=I$ (say)
$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x=I$ (say)
Solution:
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Verified Answer
$I=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} d x$
$=\frac{1}{(a-b)} \int\left[(x+a)^{1 / 2}-(x+b)^{1 / 2}\right] d x$
$=\frac{2}{3(a-b)}\left[(x+a)^{3 / 2}-(x+b)^{3 / 2}\right]+c$
$=\frac{1}{(a-b)} \int\left[(x+a)^{1 / 2}-(x+b)^{1 / 2}\right] d x$
$=\frac{2}{3(a-b)}\left[(x+a)^{3 / 2}-(x+b)^{3 / 2}\right]+c$
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