Question

`\rm(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) \\`​

Answer 1

Applying the product rule gives

`\displaystyle (d)/(dx)\int_1^(x^2)(2t)/(1+t^2)\,dt * \int_1^(\ln(x))(dt)/((1+t)^2) + \int_1^(x^2)(2t)/(1+t^2)\,dt * (d)/(dx)\int_1^(\ln(x))(dt)/((1+t)^2)`

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

`\displaystyle (4x^3)/(1+x^4) \int_1^(\ln(x))(dt)/((1+t)^2) + (1)/(x(1+\ln(x))^2)\int_1^(x^2)(2t)/(1+t^2)\,dt`

The remaining integrals are

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

`\displaystyle \int_1^(x^2)(2t)/(1+t^2)\,dt=\int_1^(x^2)(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 (4x^3)/(1+x^4) \left(\frac12-\frac1{1+\ln(x)}\right) + (1)/(x(1+\ln(x))^2) \ln\left(\frac{1+x^4}2\right)`

which could be simplified further.