int {left( {frac{{{x^3} + 2}}{{{x^3}}}} right | vedclass

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Question

(A) We are given the integral $\int {\left( {\frac{{{x^3} + 2}}{{{x^3}}}} \right)dx}$.
First,simplify the integrand by dividing each term in the nume

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We are given the integral $\int {\left( {\frac{{{x^3} + 2}}{{{x^3}}}} ight)dx}$.
First,simplify the integrand by dividing each term in the numerator by the denominator:
$\frac{{{x^3} + 2}}{{{x^3}}} = \frac{{{x^3}}}{{{x^3}}} + \frac{2}{{{x^3}}} = 1 + 2{x^{ - 3}}$.
Now,integrate the expression term by term with respect to $x$:
$\int {(1 + 2{x^{ - 3}})dx} = \int {1dx} + 2\int {{x^{ - 3}}dx}$.
Using the power rule $\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + c$ (for $n 
eq -1$):
$= x + 2\left( {\frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} ight) + c$
$= x + 2\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} ight) + c$
$= x - {x^{ - 2}} + c$
$= x - \frac{1}{{{x^2}}} + c$.

$\int {\left( {\frac{{{x^3} + 2}}{{{x^3}}}} \right)dx} $ is equal to

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