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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:

$\int\frac{1}{\sqrt{7-3\text{x}-2\text{x}^2}}\text{ dx}$

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Answer

$\int\frac{\text{dx}}{\sqrt{7-3\text{x}-2\text{x}^2}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-\frac{3}{2}\text{x}-\text{x}^2}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-\big(\text{x}^2-\frac{3}{2}\text{x}\big)}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt2}\Big)^2-\Big(\text{x}^2+\frac{3}{2}\text{x}+\big(\frac{3}{4}\big)^2-\big(\frac{3}{4}\big)^2\Big)}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt2}\Big)^2-\big(\text{x}+\frac{3}{4}\big)^2+\frac{9}{16}}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}+\frac{9}{16}-\big(\text{x}+\frac{3}{4}\big)^2}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{56+9}{16}-\big(\text{x}+\frac{3}{4}\big)^2}}$

$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt{65}}{4}\Big)^2-\big(\text{x}+\frac{3}{4}\big)^2}}$

$=\frac{1}{\sqrt2}\sin^{-1}\Bigg[\frac{\text{x}+\frac{3}{4}}{\frac{\sqrt{65}}{4}}{}\Bigg]+\text{C}$

$=\frac{1}{\sqrt2}\sin^{-1}\Big[\frac{4\text{x}+3}{\sqrt{65}}\Big]+\text{C}$

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