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

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

• $\frac{a^{x + \frac{1}{x}}}{\log_e a}$
• $\log_e a \cdot a^{x + \frac{1}{x}}$
• $\frac{a^{x + \frac{1}{x}}}{x} \log_e a$
• $x\frac{a^{x + \frac{1}{x}}}{\log_e a}$

#### 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
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#### APPEARS IN

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