Question
Evaluate the following integrals:
$\int(3\text{x}+4)^2\text{dx}$

Answer

$\int(3\text{x}+4)^2\text{dx}$
$=\int(9\text{x}^2+2\times3\text{x}\times4+16)\text{dx}$
$=9\int\text{x}^2\text{dx}+24\int\text{x dx}+16\int\text{dx}$
$=9\Big[\frac{\text{x}^3}{3}\Big]+24\Big[\frac{\text{x}^2}{2}\Big]+16\text{x}+\text{C}$
$=3\text{x}^3+12\text{x}^2+16\text{x}+\text{C}$

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