You need to use the following substitution `x^3 + 3x=u` , such that:
`x^3 + 3x = u=>(3x^2 + 3)dx = du => (x^2 + 1)dx= (du)/3`
`int (x^2 + 1)(x^3 + 3x)^4dx = (1/3)*int u^4 du`
`(1/3)*int u^4 du = (1/3)*((u^5)/5) + c`
Replacing back `x^3 + 3x` for u yields:
`int (x^2 + 1)(x^3 + 3x)^4dx = ((x^3 + 3x)^5)/15 + c`
Hence, evaluating the indefinite integral, yields `int (x^2 + 1)(x^3 + 3x)^4dx = ((x^3 + 3x)^5)/15 + c`
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