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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int(3\text{x}+1)\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
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Answer

$\text{I}=\int(3\text{x}+1)\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$Let $(3\text{x}+1)=\text{A}\frac{\text{d}}{\text{dx}}(4-3\text{x}-2\text{x}^2)+\text{B}$
$\Rightarrow(3\text{x}+1)=\text{A}(-3-4\text{x})+\text{B}$
$\Rightarrow(3\text{x}+1)=-4\text{A}\text{x}+(\text{B}-3\text{A})$
$\Rightarrow3=-4\text{A}\text{ and }(\text{B}-3\text{A})=1$
$\Rightarrow\text{A}=-\frac{3}{4}\text{ and }\text{B}=-\frac{5}{4}$
$\Rightarrow(3\text{x}+1)=-\frac{3}{4}(-3-4\text{x})-\frac{5}{4}$
$\Rightarrow\text{I}=-\frac{3}{4}(-3-4\text{x})\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}\\-\frac{5}{4}\int\sqrt{4-3\text{x}-2\text{x}^2}$
Let $\text{I}=-\frac{3}{4}\text{I}_1-\frac{5}{4}\text{I}_2\ \dots(1)$
Now,
$\text{I}_1=\int(-3-4\text{x})\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
Let $(4-3\text{x}-2\text{x}^2)=\text{t},\text{ or, }(-3-4\text{x})\text{dx = dt}$
$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$
$\Rightarrow\text{I}_1=\frac{2}{3}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}+\text{c}_1$
$\text{I}_2=\int\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
$=\int\sqrt{2\Big(2-\frac{3}{2}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\frac{17}{4}-\frac{9}{4}-\frac{3}{2}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt2\int\sqrt{\Big[\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\frac{9}{4}+\frac{3}{2}\text{x}+\text{x}^2\Big)\Big]}\text{dx}$
$=\sqrt2\int\sqrt{\Big[\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}+\frac{3}{2}\Big)^2\Big]}\text{dx}$
$=\sqrt2\sin\Bigg(\frac{\text{x}+\frac{3}{2}}{\frac{\sqrt{17}}{2}}\Bigg)+\text{c}_2$
$=\sqrt2\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{c}_2$
Using (1), we get
$\text{I}=-\frac{3}{4}\times\frac{2}{3}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}\\-\frac{5}{4}\times\sqrt2\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{C}$
$\therefore\ \text{I}=-\frac{1}{2}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}-\frac{5\sqrt2}{4}\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{C}$

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