Advertisements
Advertisements
प्रश्न
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
उत्तर
I = \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\]
\[I = \int\frac{dx}{\sqrt{- \left( x^2 + 2x - 3 \right)}}\]
\[ = \int\frac{dx}{\sqrt{- \left( x^2 + 2x - 4 + 1 \right)}}\]
\[ = \int\frac{dx}{\sqrt{- \left[ \left( x^2 + 2x + 1 \right) - 2^2 \right]}}\]
\[= \int\frac{dx}{\sqrt{- \left[ \left( x + 1 \right)^2 - 2^2 \right]}}\]
\[ = \int\frac{dx}{\sqrt{2^2 - \left( x + 1 \right)^2}}\]
\[ = \sin^{- 1} \left( \frac{x + 1}{2} \right) + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
\[\int \left( \tan x + \cot x \right)^2 dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int\frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} dx\]
\[\int\frac{1}{x (3 + \log x)} dx\]
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int \sin^5\text{ x }\text{cos x dx}\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
\[\int\frac{1}{a^2 - b^2 x^2} dx\]
\[\int\frac{1}{2 x^2 - x - 1} dx\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} \text{ dx}\]
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
\[\int \tan^3 x\ dx\]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]
