हिंदी

∫ ( 2 X + 3 X ) 2 6 X Dx - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\] 
योग
Advertisements

उत्तर

\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x}dx\]
\[ = \int\left[ \frac{\left( 2^x \right)^2 + \left( 3^x \right)^2 + 2 \cdot 2^x \cdot 3^x}{6^x} \right]dx\]
\[ = \int\left( \frac{\left( 2^x \right)^2}{2^x \cdot 3^x} + \frac{\left( 3^x \right)^2}{2^x \cdot 3^x} + \frac{2 \cdot 2^x \cdot 3^x}{2^x \cdot 3^x} \right)dx\]
\[ \Rightarrow \int\left[ \left( \frac{2}{3} \right)^x + \left( \frac{3}{2} \right)^x + 2 \right]dx\]
\[ \Rightarrow \frac{\left( \frac{2}{3} \right)^x}{\text{ ln }\left( \frac{2}{3} \right)} + \frac{\left( \frac{3}{2} \right)^x}{\text{ln } \frac{3}{2}} + 2x + C ...........\left( \because \int a^x dx = \frac{a^x}{\text{ ln } a} \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Revision Excercise | Q 6 | पृष्ठ २०३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

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

 


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

\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]

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

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

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

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

\[\int\frac{x^2}{x^6 + a^6} dx\]

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

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

\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]

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

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

\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]

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

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

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

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

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

\[\int e^x \left( \cot x + \log \sin x \right) dx\]

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

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

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

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

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

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

\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]

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

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

\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]

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

Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .


\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]


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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×