हिंदी

∫ 1 √ X + 3 − √ X + 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

Rationalising the denominator

\[= \int\frac{\left( \sqrt{x + 3} + \sqrt{x + 2} \right)}{\left( \sqrt{x + 3} - \sqrt{x + 2} \right) \left( \sqrt{x + 3} + \sqrt{x + 2} \right)} dx\]
\[ = \int\left[ \frac{\left( x + 3 \right)^\frac{1}{2} + \left( x + 2 \right)^\frac{1}{2}}{\left( x + 3 \right) - \left( x + 2 \right)} \right]dx\]
\[ = \int\left[ \left( x + 3 \right)^\frac{1}{2} + \left( x + 2 \right)^\frac{1}{2} \right]dx\]
`= [ (x+3 )^{1/2+1} / {1/2+1 }    +   (x+2)^{1/2 + 1 } / {1/2+1}] + c`
\[ = \frac{2}{3} \left( x + 3 \right)^\frac{3}{2} + \frac{2}{3} \left( x + 2 \right)^\frac{3}{2} + C\]
\[ = \frac{2}{3}\left\{ \left( x + 3 \right)^\frac{3}{2} + \left( x + 2 \right)^\frac{3}{2} \right\} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.03 | Q 17 | पृष्ठ २३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]

 
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec         } {x }- \cot x} dx\]

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

Integrate the following integrals:

\[\int\text { sin  x  cos  2x     sin 3x   dx}\]

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


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

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

\[\  ∫    x   \text{ e}^{x^2} dx\]

` ∫   tan   x   sec^4  x   dx  `


\[\int \sin^4 x \cos^3 x \text{ dx }\]

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

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

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

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

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

\[\int\frac{1}{5 + 4 \cos x} dx\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]

\[\int2 x^3 e^{x^2} dx\]

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

\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]

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

∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]

\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)} dx\]

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

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

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

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

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


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

\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]

\[\int \cot^5 x\ dx\]

\[\int \sec^6 x\ dx\]

\[\int \tan^5 x\ \sec^3 x\ dx\]

\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]

\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]


\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×