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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
$\int\frac{\text{x}}{\sqrt{\text{x}+4}}\text{dx}$
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Answer

$\int\frac{\text{x}}{\sqrt{\text{x}+4}}\text{dx}$
$=\int\Big(\frac{\text{x}+4-4}{\sqrt{\text{x}+4}}\Big)\text{dx}$
$=\int\Big(\sqrt{\text{x}+4}-\frac{4}{\sqrt{\text{x}+4}}\Big)\text{dx}$
$=\int(\text{x}+4)^\frac{1}{2}\text{dx}-4\int(\text{x}+4)^{-\frac{1}{2}}\text{dx}$
$=\frac{(\text{x}+4)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-4\frac{[\text{x}+4]^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-8(\text{x}+4)^\frac{1}{2}+\text{C}$
$=(\text{x}+4)^\frac{1}{2}\Big[\frac{2}{3}(\text{x}+4)-8\Big]+\text{C}$
$=(\text{x}+4)^\frac{1}{2}\Big[\frac{2\text{x}+8-24}{3}\Big]+\text{C}$
$=(\text{x}+4)^\frac{1}{2}\Big[\frac{2\text{x}-16}{3}\Big]+\text{C}$
$=\frac{2}{3}(\text{x}-8)\sqrt{\text{x}+4}+\text{C}$

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