मराठी

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

Advertisements
Advertisements

प्रश्न

\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]

 


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

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

`∫     cos ^4  2x   dx `


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

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

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

` ∫    \sqrt{tan x}     sec^4  x   dx `


\[\int \sec^4 2x \text{ dx }\]

Evaluate the following integrals:

\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]

Evaluate the following integrals:

\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]

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

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

\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]

\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]

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

\[\int\frac{2x - 3}{x^2 + 6x + 13} dx\]

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

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

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

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

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

\[\int \log_{10} x\ dx\]

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

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

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

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

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

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

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

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

\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]

\[\int\sqrt{\frac{a + x}{x}}dx\]
 

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

\[\int\sqrt{x^2 - a^2} \text{ dx}\]

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

Find :  \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\] 

 


\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×