Integrate `int(x^3+4)/(x^2+4)dx`

Rewrite the given function using long division.

`int[x+(-4x+4)/(x^2+4)]dx`

`=intxdx-int(4x)/(x^2+4)dx+int4/(x^2+4)dx`

Integrate the first integral using the pattern `intx^n=x^(n+1)/n+C`

`intx=x^2/2+C`

Integrate the second integral using u-substitution.

let `u=x^2+4`

`(du)/dx=2x`

`dx=(du)/(2x)`

`-int(4x)/(x^2+4)dx`

`=-4intx/u*(du)/(2x)`

` ` `=-2ln|x^2+4|+C`

`=-2ln(x^2+4)+C`

Integrate the third integral using the pattern

`int(dx)/(x^2+a^2)=(1/a)tan^-1(x/a)+C`

`int4/(x^2+4)dx=(4)(1/2)tan^-1(x/2)+C=2tan^-1(x/2)+C`

**The final answer is: **

`1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+C`

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