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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Multiple Choice Questions Question 24 Maths Textbook Solution.

Answers (1)

Answer:

\frac{a^{x+\frac{1}{x}}}{\log _{e} a}

Given:

f(x)=\left(1-\frac{1}{x^{2}}\right) a^{x+\frac{1}{x}}, a>0

Hint:

You must know about the \int a^{x} d x.

Explanation:

Let \mathrm{I}=\int\left(1-\frac{1}{x^{2}}\right) a^{x+\frac{1}{x}} d x    

       =\int a^{t} d t                                                                    \text { [Put } \left.x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^{2}}\right) d x=d t\right]

       \begin{aligned} &=\frac{a^{t}}{\log _{e} a} \\ &=\frac{a^{x+\frac{1}{x}}}{\log _{e} a} \end{aligned}                                                                        \left[\because \int a^{x} d x=\frac{a^{x}}{\log a}\right]

Hence primitive of  f(x) \text { is } \frac{a^{x+\frac{1}{x}}}{\log _{e} a}                                            

 

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