JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int\left(\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right) \log _2 x d x\) is equal to
- A \(\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x+C\)
- B \(\left(\frac{x}{2}\right)^x-\left(\frac{2}{x}\right)^x+C\)
- C \(\left(\frac{x}{2}\right)^x \log _2\left(\frac{x}{2}\right)+C\)
- D \(\left(\frac{ x }{2}\right)^{ x } \log _2\left(\frac{2}{ x }\right)+ C\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{x}{2}\right)^x-\left(\frac{2}{x}\right)^x+C\)
Step-by-step Solution
Detailed explanation
\(\int\left(x^x 2^{-x}+2^x x^{-x}\right) \log _2^x d x\) \(\int\left(e^{x \ln x} \cdot e^{-x \ln 2}+e^{x \ln 2} \cdot e^{-x \ln x}\right) d x\) \(\int\left(e^{x \ln x-x \ln 2}+e^{x \ln 2-x \ln x}\right) \frac{\ln x}{\ln 2} d x\) \(\text { let } \quad x \ln x-x \ln 2=t\)…
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