Before integrating, let's calculate: $\frac{3x^4-x^3+6x^2}{x^4}$

Let's factorize in $ℝ$: $3x^4-x^3+6x^2$

You can write the expression in factored form: $x^2·(3x^2-x+6)$

We divide the numerator and denominator of $\frac{x^2}{x^4}$ by the common factor: $x^2$, which gives the new quotient: $\frac{1}{x^2}$

$\frac{3x^2-x+6}{x^2}$

Let's integrate: $\int \frac{3x^2-x+6}{x^2}dx$

For $x$ ≠ 0, divide each numerator term by the power in the denominator: $\frac{3x^2-x+6}{x^2}=3-\frac{1}{x}+\frac{6}{x^2}$

ⅰ $\int \frac{3x^2-x+6}{x^2}dx$ = $\int 3dx-\int \frac{1}{x}dx+\int \frac{6}{x^2}dx$

$\int 3dx-\int \frac{1}{x}dx+\int \frac{6}{x^2}dx$

ⓘ The derivative of $3x$ is $3$, so $3x$ is an antiderivative of $3$: $\int 3dx$ = $3x$■The derivative of $-\ln \left|x\right|$ is $-\frac{1}{x}$, so $-\ln \left|x\right|$ is an antiderivative of $-\frac{1}{x}$: $-\int \frac{1}{x}dx$ = $-\ln \left|x\right|$■A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1: $\int \frac{6}{x^2}dx$ = $\int 6x^{-2}dx$ = $-6x^{-1}$ = $-\frac{6}{x}$

$3x-\ln \left|x\right|-\frac{6}{x}$

▶Indefinite integrals are defined up to an additive constant C, so this yields: $3x$$-\frac{6}{x}$$-\ln \left|x\right|$$+C$