हिंदी

∫ √ X ( 3 − 5 X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

योग

उत्तर

\[\int\sqrt{x}\left( 3 - 5x \right)dx\]
\[ = \int x^\frac{1}{2} \left( 3 - 5x \right)dx\]
\[ = \int\left( 3 x^\frac{1}{2} - 5 x^\frac{3}{2} \right)dx\]
\[ = 3\left[ \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] - 5\left[ \frac{x^\frac{3}{2} + 1}{\frac{3}{2} + 1} \right] + C\]
\[ = 2 x^\frac{3}{2} - 2 x^\frac{5}{2} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.02 | Q 15 | पृष्ठ १५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

\[\int \left( \tan x + \cot x \right)^2 dx\]

\[\int\frac{x^2 + x + 5}{3x + 2} dx\]

\[\int \sin^2 \frac{x}{2} dx\]

` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


\[\int\frac{1}{x (3 + \log x)} dx\]

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


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

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

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

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

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

\[\int \cos^7 x \text{ dx  } \]

\[\int\frac{1}{4 x^2 + 12x + 5} dx\]

\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]

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

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

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

`int 1/(cos x - sin x)dx`

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

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

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

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

\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]

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

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

\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to

\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]

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

\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to


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

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

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

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

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


\[\int {cosec}^4 2x\ dx\]


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

\[\int\sqrt{a^2 - x^2}\text{  dx }\]

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

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

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×