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प्रश्न
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उत्तर
\[\int\frac{dx}{\sqrt{8 + 3x - x^2}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x^2 - 3x \right)}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x^2 - 3x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 \right)}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x - \frac{3}{2} \right)^2 + \frac{9}{4}}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{\left( \frac{\sqrt{41}}{2} \right)^2 - \left( x - \frac{3}{2} \right)^2}}\]
\[ \Rightarrow \sin^{- 1} \left( \frac{x - \frac{3}{2}}{\frac{\sqrt{41}}{2}} \right) + C\]
\[ \Rightarrow \sin^{- 1} \left( \frac{2x - 3}{\sqrt{41}} \right) + C\]
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संबंधित प्रश्न
Evaluate the following integrals:
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
