Evaluate the idefinite integral integrate of (x^4+1)(sin(x))dx

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You need to use the property of linearity of integrals such that:

`int (x^4+1)sin x dx = int x^4 sin x dx + int sin x dx`

You should evaluate the integral `int x^4 sin x dx`  using integration by parts such that:

`int udv = uv - int vdu`

`u = x^4 => du = 4x^3 dx`

`dv = sin x dx => v = -cos x ` 

`int x^4 sin x dx = -x^4 cos x +4 int x^3 cos x dx`

You need to use parts to solve `int x^3 cos x dx`  such that:

`u = x^3 => du = 3x^2 dx`

`dv = cos xdx => v = sin x`

`int x^3 cos x dx = x^3 sin x - 3int x^2 sin x dx`

`int x^4 sin x dx = -x^4 cos x +4(x^3 sin x - 3int x^2 sin x dx)`

You need to solve `int x^2 sin x dx`  using parts such that:

`u = x^2 => du = 2xdx`

`dv = sin x dx => v = - cos x`

`int x^2 sin x dx = -x^2 cos x + 2 int x cos x dx`

`int x^4 sin x dx = -x^4 cos x +4(x^3 sin x - 3(-x^2 cos x + 2 int x cos x dx))`

You need to integrate `int x cos x dx`  using parts such that:

`u = x => du = dx`

`dv = cos x dx => v = sin x`

`int x cos x dx = x sin x - int sin x dx`

`int x cos x dx = x sin x + cos x + c`

`int x^4 sin x dx = -x^4 cos x +4(x^3 sin x - 3(-x^2 cos x + 2(x sin x + cos x))) + c`

`int x^4 sin x dx = -x^4 cos x +4(x^3 sin x - 3(-x^2 cos x + 2xsin x + 2 cos x)) + c`

`int x^4 sin x dx = -x^4 cos x + 4(x^3 sin x + 3x^2 cos x - 6x sin x - 6 cos x) + c`

`int x^4 sin x dx = -x^4 cos x + 4x^3 sin x + 12 x^2 cos x - 24 x sin x - 24 cos x + c`

`int (x^4+1)sin x dx = -x^4 cos x + 4x^3 sin x + 12 x^2 cos x - 24 x sin x - 24 cos x - cos x + c`

`int (x^4+1)sin x dx = -x^4 cos x + 4x^3 sin x + 12 x^2 cos x - 24 x sin x - 25 cos x + c`

Hence, evaluating the given integral, under the given conditions, yields `int (x^4+1)sin x dx = -x^4 cos x + 4x^3 sin x + 12 x^2 cos x - 24 x sin x - 25 cos x + c.`

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