Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
योग
उत्तर
\[\text{ We have, } \]
\[I = \int \frac{dx}{\left( x - 1 \right) \sqrt{x + 2}}\]
\[\text{ Putting x} + 2 = t^2 \]
\[ \Rightarrow dx = 2t \text{ dt}\]
\[ \therefore I = \int\frac{2t \text{ dt}}{\left( t^2 - 2 - 1 \right)t}\]
\[ = \int \frac{2 \text{ dt }}{t^2 - \left( \sqrt{3} \right)^2}\]
\[ = 2 \times \frac{1}{2\sqrt{3}}\text{ log }\left| \frac{t - \sqrt{3}}{t + \sqrt{3}} \right| + C\]
\[ = \frac{1}{\sqrt{3}}\text{ log }\left| \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} \right| + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
` ∫ cos 3x cos 4x` dx
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
\[\int\frac{1}{x (3 + \log x)} dx\]
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]
\[\int \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{ dx }\]
\[\int x e^x \text{ dx }\]
\[\int2 x^3 e^{x^2} dx\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int\sqrt{3 - x^2} \text{ dx}\]
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
\[\int\frac{1}{x^4 - 1} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int \log_{10} x\ dx\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
