Question

$\int\left(\frac{1}{x^{2}}+\frac{6}{x^{3}}\right) d x$

Answer 1

$ \textbf{Answer:}$

$ -\frac{1}{x}- \frac{3}{x^2}+C$

$\textbf{Explaination:}$

$ \int \left( \frac{1}{x^2} + \frac{6}{x^3} \right) dx =  \int \frac{1}{x^2} dx + \int \frac {6}{x^3}dx$  (sum rule of integrals)

$ = \int x^{-2} dx + 6 \int x^{-3} dx$  (writing the fractions in terms of the exponents and taking out constants out of the integral)

$ = \frac{1}{-2+1}x^{-2+1} + c_1 + \frac{6}{-3+1}x^{-3+1} + c_2$  (integration exponents rule)

$ = -\frac{1}{x} + c_1 -\frac{3}{x^2} + c_2$  (simplifying)

$ = -\frac{1}{x} -\frac{3}{x^2} +C $   ($ C = c_1 + c_2$ is a constant of integration)