Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
योग
उत्तर
\[\int\frac{dx}{\sqrt{\left( 2 - x \right)^2 - 1}}\]
\[\text{ let 2 }- x = t\]
\[ \Rightarrow - dx = dt\]
\[ \Rightarrow dx = - dt\]
\[Now, \int\frac{dx}{\sqrt{\left( 2 - x \right)^2 - 1}}\]
\[ = \int\frac{- dt}{\sqrt{t^2 - 1}}\]
\[ = - \text{ log }\left| t + \sqrt{t^2 - 1} \right| + C\]
\[ = - \text{ log }\left| \left( 2 - x \right) + \sqrt{\left( 2 - x \right)^2 - 1} \right| + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\frac{x^3}{x - 2} dx\]
`∫ cos ^4 2x dx `
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]
\[\int\frac{x + 1}{x^2 + x + 3} dx\]
\[\int x e^{2x} \text{ dx }\]
\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2} \text{ dx }\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]
\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]
\[\int\frac{x^3 - 1}{x^3 + x} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]
If \[\int\frac{2^{1/x}}{x^2} dx = k 2^{1/x} + C,\] then k is equal to
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
