Q.Evaluate $\displaystyle\int \left(3x^{4} - \frac{2}{x} + 5e^{x}\right)dx$.

Linearity lets each term be integrated separately, applying the power-rule, the $1/x$ log-rule, and the exponential rule in turn.

$\displaystyle\int 3x^{4}dx=\dfrac{3x^{5}}{5}$; $\displaystyle\int\frac{-2}{x}dx=-2\ln|x|$; $\displaystyle\int5e^{x}dx=5e^{x}$. Add one constant $C$.

By linearity, integrate each term separately using standard formulas 1, 2 and 3:

Term 1: $\displaystyle\int 3x^{4}\,dx = 3\cdot\frac{x^{5}}{5}=\frac{3x^{5}}{5}$ (formula 1, $n=4$).

Term 2: $\displaystyle\int -\frac{2}{x}\,dx = -2\ln|x|$ (formula 2).

Term 3: $\displaystyle\int 5e^{x}\,dx = 5e^{x}$ (formula 3).

Combining, with a single constant of integration for the whole expression:

Check by differentiation: $\dfrac{d}{dx}\left[\dfrac{3x^{5}}{5}-2\ln|x|+5e^{x}\right] = 3x^{4}-\dfrac{2}{x}+5e^{x}$, which matches the original integrand exactly.

Writing $\int\frac{2}{x}dx$ as $\frac{2x^{0}}{0}$ by mistakenly applying the power-rule formula instead of the separate $\ln|x|$ rule — formula 1 explicitly excludes $n=-1$, exactly because $1/x=x^{-1}$ needs formula 2 instead. Also, writing a separate $+C$ after every term instead of one combined constant at the end.