A general rule for integration states that the integral of a term x^n is given as x^(n+1) / (n+1) for x not equal to -1. Also the integral of 1/ x is ln x.

Now y = (2x^3 + 3x + 1) / x^4

=> Y = (2x^3/ x^4) + (3x/ x^4) + (1/ x^4)

=> Y= (2/ x) + (3 / x^3) + (1/ x^4)

=> Y = 2/ x + 3* x^ (-3) + x^(-4)

The integral with respect to x is:

2*ln x + 3*x^(-3+1) / (-3+1) + x^(-4+1) / (-4+1)

=> 2*ln x – 3*x^-2/ (2) - x^(-3) / 3

=> 2* ln x – (3/ 2x^2) – 1/ (3*x^3)

**Therefore the result is 2* ln x – 3 / 2x^2 – 1 / (3*x^3)**

To integrate a function f(x) means to determine the function F(x), which, when is differentiated, gives back the function f(x).

We'll integrate f(x) = y.

Int f(x)dx =Int (2x^3+3x+1) dx/x^4

We'll apply the linear property of integrals and we'll get:

Int (2x^3+3x+1) dx/x^4 = Int 2x^3dx/x^4 + Int 3xdx/x^4 + Int dx/x^4

We'll calculate each Integral:

We'll start with Int 2x^3dx/x^4

We'll reduce like terms:

Int 2x^3dx/x^4 = Int 2dx/x

Int 2dx/x = 2Int dx/x

**2Int dx/x = 2ln x + C**

Now, we'll calculate Int 3xdx/x^4.

Int 3xdx/x^4 = Int 3dx/x^3

Int 3dx/x^3 = 3Int x^-3*dx

3Int x^-3*dx = 3*x^(-3+1)/(-3+1) + C

**3*x^(-3+1)/(-3+1) + C = -3/2x^2 + C**

Finally, we'll calculate Int dx/x^4:

Int dx/x^4 = Int x^-4*dx

Int x^-4*dx = x^(-4+1)/(-4+1) + C

x^(-4+1)/(-4+1) + C = -1/3x^3 + C

Now, the Integrals is:

Int f(x)dx = 2ln x - 3/2x^2 - 1/3x^3 + C

**Int f(x)dx = ln x^2 - 3/2x^2 - 1/3x^3 + C**

To integrate y = (2x^3+3x+1)/x^4

Solution:

We know Int kx^n dx = k{x^(n+1)}/(n+1) +C.

Also int 1/x dx = lnx +const.

So Int y dx = {(2x^3+3x+1)/x^4} dx = Int (x^3/x^4+x/x^4+1/x^4}dx

Int ydx = 2Int /x dx + 3Int 1/x^3 dx + Int 1/x^4 dx

Int ydx = 2lnx + 3*{x^(-3+1)}/(-3+1) + x^(-4+1)/(-4+1) +ct

Int ydx = 2lnx - 3/2x ^2 -1/3x^3 +const

Int ydx = Int