Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
योग
Advertisements
उत्तर
\[\text{ Let I} = \int\frac{1}{\sqrt{3 - 2x - x^2}}dx\]
\[ = \int\frac{1}{\sqrt{3 - \left( x^2 + 2x + 1 - 1 \right)}}dx\]
\[ = \int\frac{1}{\sqrt{4 - \left( x + 1 \right)^2}}dx\]
\[\text{ Putting} \left( x + 1 \right) = t\]
\[ \Rightarrow dx = dt\]
\[ \therefore I = \int\frac{dt}{\sqrt{2^2 - t^2}}\]
\[ = \sin^{- 1} \left( \frac{t}{2} \right) + C .................\left[ \because \int \frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \sin^{- 1} \left( \frac{x + 1}{2} \right) + C .....................\left[ \because t = \left( x + 1 \right) \right]\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]
\[\int \cos^2 \frac{x}{2} dx\]
` ∫ sin 4x cos 7x dx `
` ∫ cos 3x cos 4x` dx
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
` ∫ tan 2x tan 3x tan 5x dx `
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]
\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]
\[\int\frac{1}{\cos 2x + 3 \sin^2 x} dx\]
`int 1/(cos x - sin x)dx`
`int"x"^"n"."log" "x" "dx"`
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
\[\int\left( x + 1 \right) \sqrt{2 x^2 + 3} \text{ dx}\]
\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int\frac{\log \left( 1 - x \right)}{x^2} \text{ dx}\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
