Question

Find the integral of the given function w.r.t - x

$y=x-\frac{1}{x}+\frac{1}{x^2}$

• A. $\frac{x^2}{2}-lnx+\frac{1}{x}+c$
• B. $\frac{x^2}{2}+lnx+\frac{1}{x}+c$
• C. $\frac{x^2}{2}-lnx-\frac{1}{x}+c$
• D. None of these

Answer 1

The correct option is C

$\frac{x^2}{2}-lnx-\frac{1}{x}+c$

$\int ydx=\int xdx-\int \frac{1}{x}dx+\int \frac{1}{x^2}dx$

= $\frac{x^2}{2}-lnx-\frac{x^{-2+1}}{-2+1}+c$

= $\frac{x^2}{2}-lnx-\frac{1}{x}+c$