English

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

Advertisements
Advertisements

Question

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

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Revision Excercise [Page 203]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Revision Excercise | Q 6 | Page 203

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


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

`∫     cos ^4  2x   dx `


` ∫  tan 2x tan 3x  tan 5x    dx  `

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

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

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

Evaluate the following integrals:

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

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

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

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

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

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

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

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

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

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

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

\[\int x \cos x\ dx\]

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

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

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

\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]

\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]

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

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

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

\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]

\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]

 


Write a value of

\[\int e^{3 \text{ log x}} x^4\text{ dx}\]

\[\int\left( x - 1 \right) e^{- x} dx\] is equal to

\[\int \cos^3 (3x)\ dx\]

\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]

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

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×