Question

integration of this question

Answer 1

Answer :

Now,

$\int \frac{ {x}^{3} }{x + 2} dx \\ \\ = \int \frac{( {x}^{3} + 8) - 8 }{x + 2} dx \\ \\ = \int \frac{(x + 2)( {x}^{2} - 2x + 4) - 8}{x + 2} dx \\ \\ = \int( {x}^{2} - 2x + 4)dx - 8 \int \frac{dx}{x + 2} \\ \\ = \int {x}^{2} dx - 2 \int xdx + 4 \int dx - 8 \int \frac{dx}{x + 2} \\ \\ = \frac{1}{3} {x}^{3} - {x}^{2} + 4x - 8 \log |x + 2| + c$

where c is integral constant

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Answer 2

Answer:

Step-by-step explanation: