Question

x%7D%5E%7B2%7D%20%7B%7D%7B%7D%20%20%7D%20%20%5Cfrac%7B2t%7D%7B1%20%2B%20%20%7B%20t%7D%5E%7B2%7D%20%7D%20dt%5Cbigg%29%20%5Cbigg%28%20%5Cint_%7B%201%20%7D%5E%7B%20lnx%7D%20%20%5Cfrac%7B1%7D%7B%281%20%2B%20%20%7Bt%29%7D%5E%7B2%7D%20%7Ddt%20%5Cbigg%29%5Cright%29%20%5C%5C%20" id="TexFormula1" title=" \rm\frac{d}{dx} \left ( \bigg( \int_{1}^{ {x}^{2} {}{} } \frac{2t}{1 + { t}^{2} } dt\bigg) \bigg( \int_{ 1 }^{ lnx} \frac{1}{(1 + {t)}^{2} }dt \bigg)\right) \\ " alt=" \rm\frac{d}{dx} \left ( \bigg( \int_{1}^{ {x}^{2} {}{} } \frac{2t}{1 + { t}^{2} } dt\bigg) \bigg( \int_{ 1 }^{ lnx} \frac{1}{(1 + {t)}^{2} }dt \bigg)\right) \\ " align="absmiddle" class="latex-formula">​

Answer 1

Applying the product rule gives

$\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}$

Use the fundamental theorem of calculus to compute the remaining derivatives.

$\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt$

The remaining integrals are

$\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}$

$\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)$

and so the overall derivative is

$\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)$

which could be simplified further.