हिंदी

∫ 1 ( X − 1 ) √ X + 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
योग
Advertisements

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?
अध्याय 19: Indefinite Integrals - Exercise 19.32 [पृष्ठ १९६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.32 | Q 1 | पृष्ठ १९६

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

`int{sqrtx(ax^2+bx+c)}dx`

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


\[\int\frac{1}{1 + \cos 2x} dx\]

\[\int\frac{1}{1 - \cos 2x} dx\]

\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2}  \text{dx} \]

\[\int\sin x\sqrt{1 + \cos 2x} dx\]

\[\int\frac{1 + \cos x}{1 - \cos x} dx\]

\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]

\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]

\[\int \sin^2\text{ b x dx}\]

\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]

` ∫  {sin 2x} /{a cos^2  x  + b sin^2  x }  ` dx 


` ∫ {"cosec"   x }/ { log  tan   x/2 ` dx 

\[\int \sin^5\text{ x }\text{cos x dx}\]

\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]

\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]

\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]

\[\int\frac{1}{a^2 - b^2 x^2} dx\]

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]

\[\int x^2 \text{ cos x dx }\]

\[\int\frac{\log x}{x^n}\text{  dx }\]

\[\int x \sin^3 x\ dx\]

\[\int \cos^3 \sqrt{x}\ dx\]

\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]

\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]

\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]

\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]

Evaluate the following integral:

\[\int\frac{x^2}{1 - x^4}dx\]

\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 


\[\int \sec^2 x \cos^2 2x \text{ dx }\]

\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]

\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]

\[\int\frac{x + 1}{x^2 + 4x + 5} \text{  dx}\]

\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]


\[\int \sec^{- 1} \sqrt{x}\ dx\]

\[\int \left( \sin^{- 1} x \right)^3 dx\]

\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
 

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×