Advertisements
Advertisements
प्रश्न
उत्तर
\[\int\frac{dx}{\sqrt{5 - 4x - 2 x^2}}\]
\[ = \int\frac{dx}{\sqrt{2\left[ \frac{5}{2} - 2x - x^2 \right]}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - 2x - x^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x^2 + 2x \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x^2 + 2x + 1 - 1 \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x + 1 \right)^2 + 1}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \left( x + 1 \right)^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x + 1 \right)^2}}\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left( \frac{\left( x + 1 \right)\sqrt{2}}{\sqrt{7}} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left( \sqrt{\frac{2}{7}}\left( x + 1 \right) \right) + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
