# ∫ 1 √ 7 − 3 X − 2 X 2 D X - Mathematics

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

#### Solution

$\int\frac{dx}{\sqrt{7 - 3x - 2 x^2}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \frac{3}{2}x - x^2}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \left( x^2 - \frac{3}{2}x \right)}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x^2 + \frac{3}{2}x + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 \right)}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x + \frac{3}{4} \right)^2 + \frac{9}{16}}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} + \frac{9}{16} - \left( x + \frac{3}{4} \right)^2}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{56 + 9}{16} - \left( x + \frac{3}{4} \right)^2}}$
$= \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{65}}{4} \right)^2 - \left( x + \frac{3}{4} \right)^2}}$
$= \frac{1}{\sqrt{2}} \sin^{- 1} \left[ \frac{x + \frac{3}{4}}{\frac{\sqrt{65}}{4}} \right] + C$
$= \frac{1}{\sqrt{2}} \sin^{- 1} \left[ \frac{4x + 3}{\sqrt{65}} \right] + C$

Concept: Indefinite Integral Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.17 | Q 6 | Page 93