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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 10 Maths Textbook Solution.

Answers (1)

Answer:

-\frac{1}{\log _{e} 2}

Given:

\int \frac{2^{\frac{1}{x}}}{x^{2}} d x=K 2^{\frac{1}{x}}+C

Hint

Using  \int a^{x} d x

Explanation:

Let =\int \frac{2^{\frac{1}{x}}}{x^{2}} d x                                    \text { [Put } \frac{1}{x}=t \Rightarrow-\frac{1}{x^{2}} d x=d t \Rightarrow \frac{d x}{x^{2}}=-d t

      \begin{aligned} &=-\int 2^{t} d t \\ &=-\frac{2^{t}}{l o g_{e} 2}+2 \end{aligned}

      =-\frac{2 \frac{1}{x}}{\log _{e} 2}+2                                \left[\because \int a^{x} d x=\frac{a^{x}}{\log _{e} a}\right]

According to given:I=K 2^{\frac{1}{x}}+C

                 \therefore K 2^{\frac{1}{x}}+C=-\frac{1}{\log _{e} 2} \cdot 2^{\frac{1}{x}}+C

Comparing both sides,

               K=\frac{-1}{\log _{e} 2}

 

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