can anyone integarte and simplify 2x^3-3x/4x

Question

kropot72 · Answer

Is this the integral to be solved?
$$\int\limits_{}^{}((2x ^{3}-3x)\div(4x))dx$$

anonymous · Answer

yes

anonymous · Answer

Assuming you meant $$ \frac{  2x^3 - 3x}{4x} = \frac{2x^3}{4x} - \frac{3x}{4x}  $$, where we can integrate the two bit : 2x^3/4x has for integral x^3/6 and 3x/4x has 3x/4 (respectively from the polinomial+quotient and quotient rule ). So the final result is -((3 x)/4) + x^3/6 which can be factored as (1/12) (x) (-9 + 2 x^2)

campbell_st · Answer

simplify 1st

$$\int\limits 2x^3/4x - 3x/4x dx = \int\limits x^2/2 - 3/4  dx$$

kropot72 · Answer

If both terms in the denominator are divided by 4x the result is:
$$\int\limits_{}^{}((x ^{2}/2)-(3/4))dx=(x ^{3}/6)-(3x/4)+C$$

campbell_st · Answer

this becomes $$1/2 \times  x^3/3  3x/4 + C = x^3/6 - 3x/4 + C$$

kropot72 · Answer

My reply should have said"If both terms in the numerator.............". Sorry for the slip:)