Advertisement Remove all ads

The Primitive of the Function F ( X ) = ( 1 − 1 X 2 ) a X + 1 X , a > 0 is - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
MCQ

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]

Options

  • \[\frac{a^{x + \frac{1}{x}}}{\log_e a}\]
  • \[\log_e a \cdot a^{x + \frac{1}{x}}\]
  • \[\frac{a^{x + \frac{1}{x}}}{x} \log_e a\]
  • \[x\frac{a^{x + \frac{1}{x}}}{\log_e a}\]
Advertisement Remove all ads

Solution

\[\frac{a^{x + \frac{1}{x}}}{\log_e a}\]
 
 
\[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) \cdot a^{x + \frac{1}{x}} \]
\[ \therefore \int f\left( x \right)dx = \int\left( 1 - \frac{1}{x^2} \right) \cdot a^{x + \frac{1}{x}} dx\]

\[\text{Let }\left( x + \frac{1}{x} \right) = t\]
\[ \Rightarrow \left( 1 - \frac{1}{x^2} \right)dx = dt\]
\[ \therefore \int f\left( x \right)dx = \int a^t \cdot dt\]
\[ = \frac{a^t}{\log_e a} + C\]
\[ = \frac{a^{x + \frac{1}{x}}}{\log_e a} + C ...........\left( \because t = x + \frac{1}{x} \right)\]

Concept: Indefinite Integral Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
MCQ | Q 24 | Page 202

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×