Integral \(\int\left(\frac{2 x^3-3 x+5}{2 x^2}\right) d x\) is valid for
Integral \(\int\left(\frac{2 x^3-3 x+5}{2 x^2}\right) d x\) is valid for
\(x \in \mathrm{R}-\{0\}\)
\(x>0\)
\(x < 0\)
\(x \in \mathrm{R}\)
Solution
\(\begin{aligned}
I & =\int\left(\frac{2 x^3-3 x+5}{2 x^2}\right) d x \\
& =\int\left(x-\frac{3}{2 x}+\frac{5}{2 x^2}\right) d x \\
& =\frac{x^2}{2}-\frac{3}{2} \log _e x-\frac{5}{2} \frac{1}{x}+c
\end{aligned}\)
is valid for \(x > 0\), because \(\log _e x\) valid, if \(x > 0\) Hence, option (b) is correct.