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Mathematics 19 Online
OpenStudy (anonymous):

fundamental theorem or calculus help? (problem in comments.)

OpenStudy (anonymous):

\[\frac{ d }{ dx }(\int\limits_{0}^{x^{4}}\ln (t^{2}+1)dt)\]

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

@agent0smith

ganeshie8 (ganeshie8):

\(\frac{ d }{ dx }(\int\limits_{0}^{x^{4}}\ln (t^{2}+1)dt) \\ = \dfrac{d}{dx}\left[F(x^4) - F(0)\right] \\= F'(x^4) (x^4)' - 0 \\ = \ln((x^4)^2 + 1) * 4x^3\)

ganeshie8 (ganeshie8):

btw, your \(F'(t) = \ln(t^2+1)\) \(\implies F(t) = \int F'(t) dt \)

OpenStudy (anonymous):

Ah, I see! thanks for your help.

ganeshie8 (ganeshie8):

In short : just apply the FTC, and apply chain rule for the function in bounds

ganeshie8 (ganeshie8):

\(\frac{ d }{ dx }(\int\limits_{0}^{x^{4}}\ln (t^{2}+1)dt) = \ln((x^4)^2 + 1) * (x^4)' \)

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