## Related questions with answers

Carry out the differentiation.

$\displaystyle \frac{d}{dx}\left(\int_x^{x^2} \frac{dt}{1+t^2}\right)$.

Solution

Verified$\begin{gathered} \frac{d}{{dx}}\left( {\int_x^{{x^2}} {\frac{{dt}}{{1 + {t^2}}}} } \right) \\ \textcolor{#4257b2}{ {\text{Apply }}\frac{d}{{dx}}\left( {\int_{u\left( x \right)}^{v\left( x \right)} {f\left( t \right)dt} } \right) = f\left( {v\left( x \right)} \right)v'\left( x \right) - f\left( {u\left( x \right)} \right)u'\left( x \right)} \\ {\text{Let }}f\left( t \right) = \frac{1}{{1 + {t^2}}},{\text{ }}v\left( x \right) = {x^2},{\text{ and }}u\left( x \right) = x \\ \textcolor{#4257b2}{{\text{Therefore}}{\text{,}}} \\ \frac{d}{{dx}}\left( {\int_x^{{x^2}} {\frac{{dt}}{{1 + {t^2}}}} } \right) = \underbrace {\left[ {\frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}}} \right]}_{f\left( {v\left( x \right)} \right)}\underbrace {\frac{d}{{dx}}\left[ {{x^2}} \right]}_{v'\left( x \right)} - \underbrace {\left[ {\frac{1}{{1 + {{\left( x \right)}^2}}}} \right]}_{f\left( {u\left( x \right)} \right)}\underbrace {\frac{d}{{dx}}\left[ x \right]}_{u'\left( x \right)} \\ \textcolor{#4257b2}{ {\text{Differentiate and simplify}}} \\ \frac{d}{{dx}}\left( {\int_x^{{x^2}} {\frac{{dt}}{{1 + {t^2}}}} } \right) = \left( {\frac{1}{{1 + {x^4}}}} \right)\left( {2x} \right) - \left( {\frac{1}{{1 + {x^2}}}} \right)\left( 1 \right) \\ \frac{d}{{dx}}\left( {\int_x^{{x^2}} {\frac{{dt}}{{1 + {t^2}}}} } \right) = \frac{{2x}}{{1 + {x^4}}} - \frac{1}{{1 + {x^2}}} \\ \frac{d}{{dx}}\left( {\int_x^{{x^2}} {\frac{{dt}}{{1 + {t^2}}}} } \right) = \frac{{2x + 2{x^3} - 1 - {x^4}}}{{\left( {1 + {x^4}} \right)\left( {1 + {x^2}} \right)}} \\ \end{gathered}$

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