Problem

Find the antiderivative of the given derivative.
\[
\frac{d s}{d t}=12 t\left(4 t^{2}-7\right)^{3}
\]
\[
s=\square
\]

Answer

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Answer

So, the antiderivative of the given derivative is \(\boxed{s = (3/16)(4t^2 - 7)^4 + C}\)

Steps

Step 1 :Given the derivative \(\frac{d s}{d t}=12 t\left(4 t^{2}-7\right)^{3}\)

Step 2 :Recognize this as a case of the chain rule for derivatives, where the outer function is \((u)^3\) and the inner function is \(4t^2 - 7\), with its derivative being \(8t\)

Step 3 :Use the method of substitution. Let's set \(u = 4t^2 - 7\). Then, \(du = 8t dt\)

Step 4 :Rewrite 12 as 3*4, and then the integral becomes \(\int 3*4t\left(4 t^{2}-7\right)^{3} dt\)

Step 5 :Substitute \(u\) and \(du\) into the integral: \(\int 3u^3 * (1/4) du\)

Step 6 :Solve this integral to get \(s = (3/4) * (1/4)u^4 + C\)

Step 7 :Substitute back for \(u\) to get \(s = (3/16)(4t^2 - 7)^4 + C\)

Step 8 :So, the antiderivative of the given derivative is \(\boxed{s = (3/16)(4t^2 - 7)^4 + C}\)

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