Question

Which of the following is an antiderivative of f(x)=x3e4x? Hint: Do not integrate directly. Differentiate and check.

Answer 1

If we do not integrate directly, we need to differentiate the answers and check, so let's start:

a)

$(\frac{1}{16}x^4e^{4x}-1)^{\prime}=\frac{1}{16}(4x^3e^{4x}+x^4e^{4x}4)=e^{4x}x^3\frac{4}{16}(1+x)$

So we discard this option.

b)

$(\frac{1}{128}e^{4x}(32x^3-24x^2+12x-3)+10)^{\prime}=\frac{1}{128}(4e^{4x}(32x^3-24x^2+12x-3)+e^{4x}(96x^2-48x+12))$

If we simply:

$\frac{1}{128}e^{4x}(128x^3-96x^2+48x-12+96x^2-48x+12)$

This is equal to:

$e^{4x}x^3$

Then the correct answer is B.