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\(\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=\)
Options:
Solution:
2207 Upvotes
Verified Answer
The correct answer is:
\(\frac{x}{\log x}+c\)
\(\int\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right) d x\)
Put, \(\log x=t\)
\(\begin{array}{l}
x=e^t \\
d x=e^t d t \\
\int\left(\frac{1}{t}+\frac{-1}{t^2}\right) e^t d t
\end{array}\)
We have,
\(\begin{aligned}
\int e^x\left(f(x)+f^{\prime}(x) d x\right. & \left.=e^x \cdot f(x)+C\right) \\
& =e^t \cdot \frac{1}{t}+c=\frac{e^{\log x}}{\log x}+c=\frac{x}{\log x}+c
\end{aligned}\)
Hence, option (d) is correct.
Put, \(\log x=t\)
\(\begin{array}{l}
x=e^t \\
d x=e^t d t \\
\int\left(\frac{1}{t}+\frac{-1}{t^2}\right) e^t d t
\end{array}\)
We have,
\(\begin{aligned}
\int e^x\left(f(x)+f^{\prime}(x) d x\right. & \left.=e^x \cdot f(x)+C\right) \\
& =e^t \cdot \frac{1}{t}+c=\frac{e^{\log x}}{\log x}+c=\frac{x}{\log x}+c
\end{aligned}\)
Hence, option (d) is correct.
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