Jambmaths question:
$\begin{align} & \text{Integrate }\left( \frac{1+x}{{{x}^{3}}} \right)dx \\ & \text{(A) }-\frac{1}{2{{x}^{2}}}-\frac{1}{x}+k\text{ (B) }-\frac{{{x}^{2}}}{2}-\frac{1}{x}+k\text{ (C) }{{x}^{2}}-\frac{1}{x}+k\text{ (D) }2{{x}^{2}}-\frac{1}{x}+k \\\end{align}$
Jamb Maths Solution:
$\begin{align} & \int{\frac{1+x}{{{x}^{3}}}dx}=\int{\left( \frac{1}{{{x}^{3}}}+\frac{x}{{{x}^{3}}} \right)dx}=\int{({{x}^{-3}}+{{x}^{-2}})dx} \\ & \int{\frac{1+x}{{{x}^{2}}}dx}=\frac{{{x}^{-2}}}{-2}+\frac{{{x}^{-1}}}{-1}+K \\ & \int{\frac{1+x}{{{x}^{2}}}dx}=-\frac{1}{2{{x}^{2}}}-\frac{1}{x}+K \\\end{align}$
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