मराठी

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

Advertisements
Advertisements

प्रश्न

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

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

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

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

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


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

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

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

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

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

` ∫  tan 2x tan 3x  tan 5x    dx  `

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

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

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

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

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

\[\int\frac{1}{x^2 - 10x + 34} dx\]

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

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

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

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

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

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

\[\int x^2 \sin^2 x\ dx\]

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

\[\int e^x \sec x \left( 1 + \tan x \right) dx\]

\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]

\[\int\sqrt{3 - x^2} \text{ dx}\]

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

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

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

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

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

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


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

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

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

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


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

\[\int\sqrt{3 x^2 + 4x + 1}\text{  dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×