हिंदी

∫ ( 3 X + 4 ) 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \left( 3x + 4 \right)^2 dx\]
योग
Advertisements

उत्तर

\[\int \left( 3x + 4 \right)^2 dx\]
\[ = \int \left( 9 x^2 + 2 \times 3x \times 4 + 16 \right)dx\]
`= 9 ∫    x^2dx + 24  ∫  x dx + 16 ∫   dx`
\[ = 9\left[ \frac{x^3}{3} \right] + 24\left[ \frac{x^2}{2} \right] + 16x + C\]
\[ = 3 x^3 + 12 x^2 + 16x + C\]

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 18 | पृष्ठ १५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

 


\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]

` ∫  1/ {1+ cos   3x}  ` dx


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

\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]

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

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

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

\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]

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

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

Evaluate the following integrals:

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

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

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

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

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

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

`int 1/(sin x - sqrt3 cos x) dx`

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

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

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

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

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

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

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

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

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

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

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

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


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

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

\[\int \sec^4 x\ dx\]


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

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


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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×