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Definite Integration (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Definite Integration (p-1)3 Marks
Question
$\int_3^5 \frac{d x}{\sqrt{x+4}+\sqrt{x-2}}$
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Answer

$
\begin{aligned}
& \int_3^5 \frac{d x}{\sqrt{x+4}+\sqrt{x-2}} \\
& =\int_3^5 \frac{1}{\sqrt{x+4}+\sqrt{x-2}} \times \frac{\sqrt{x+4}-\sqrt{x-2}}{\sqrt{x+4}-\sqrt{x-2}} d x \\
& =\int_3^5 \frac{\sqrt{x+4}-\sqrt{x-2}}{x+4-x+2} d x \\
& =\frac{1}{6} \int_3^5\left[(x+4)^{\frac{1}{2}}-(x-2)^{\frac{1}{2}}\right] d x \\
& =\frac{1}{6}\left[\frac{(x+4)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{(x-2)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_3^5 \\
& =\frac{1}{9}\left[(x+4)^{\frac{3}{2}}-(x-2)^{\frac{3}{2}}\right]_3^5 \\
& =\frac{1}{9}\left[\left(9^{\frac{3}{2}}-3^{\frac{3}{2}}\right)-\left(7^{\frac{3}{2}}-1\right)\right] \\
& =\frac{1}{9}(27-3 \sqrt{3}-7 \sqrt{7}+1) \\
& =\frac{1}{9}(28-3 \sqrt{3}-7 \sqrt{7}) .
\end{aligned}
$

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