Question

$\rm \frac{d}{dx} \bigg( \int_{ {1} }^{ {x}^{2} } \frac{2t}{1 + { t }^{2} } dt \cdot \int_{ 1}^{lnx} \frac{1}{(1 + t {)}^{2} } dt \bigg)$​

Answer 1

$\large\sf\red{\underline{\underline{Question :}}}$

• $\displaystyle\dfrac{d}{dx}\left(\int _1 ^{x^{2}} \dfrac{2t}{1+t^2}dt.\int ^{\ln x} _1 \dfrac{1}{(1+t)^2}\right)$

$\large\sf\red{\underline{\underline{Solution :}}}$

$= \displaystyle\dfrac{d}{dx}\left(\int _1 ^{x^{2}} \dfrac{2t}{1+t^2}dt.\int ^{\ln x} _1 \dfrac{1}{(1+t)^2}\right)$

✮ Integrating :

$\boxed{\red\bigstar \displaystyle\int\sf \dfrac{f(x)'}{f(x)}= \log( f(x))}\;\;\;\;\;\;\;\boxed{\red\bigstar\int x^n = \dfrac{x^{n+1}}{n+1}}$

$= \displaystyle\dfrac{d}{dy}\left( \left[\log(1+t^2)\right]^{x^2} _1 . \int ^{\ln x} _1 (1+t)^{-2} \right)$

$= \displaystyle\dfrac{d}{dy}\left( \left[\log(1+t^2)\right]^{x^2} _1 . \left[\dfrac{-1}{1+t}\right]^{\ln x} _1 \right)$

✮  Substituting limits :  

$\rm\displaystyle = \dfrac{d}{dy}\left( \left[\log\left(1+({x^2})^2\right)-\log(1+1^2)\right]. \left[\dfrac{-1}{1+\ln x}-\left(\dfrac{-1}{1+1}\right)\right] \right)$

$\rm\displaystyle=\dfrac{d}{dy}\left( \left[\log\left(1+x^4\right)-\log(2)\right]. \left[\dfrac{1}{2}-\dfrac{1}{1+\ln x}\right] \right)$

✮ Differentaiting using product rule :

• (UV)' = U'V + V'U
• $\dfrac{d}{dy}\log x = \dfrac{1}{x}$

$\rm\displaystyle= \dfrac{d}{dy}\left( \left[\log\left(1+x^4\right)-\log(2)\right]\right)'\left[\dfrac{1}{2}-\dfrac{1}{1+\ln x}\right] +\dfrac{d}{dy}\left(\dfrac{1}{2}-\dfrac{1}{1+\ln x}\right)' \;\left[\log\left(1+x^4\right)-\log(2)\right]$

$\rm\displaystyle=\left(\dfrac{1}{1+x^4}.4x^3-0\right)\left(\dfrac{1}{2}-\dfrac{1}{1+\ln x}\right) +\left(\dfrac{1}{(1+\ln x)^2}.(0+\frac{1}{x})\right)\left(\log\left(1+x^4\right)-\log(2)\right)$

$\rm\displaystyle=\dfrac{4x^3}{1+x^4}\left(\dfrac{1}{2}-\dfrac{1}{1+\ln x}\right)+\dfrac{\log(1+x^4)-\log 2 }{x(1+\ln x)^2}$

$\rm\displaystyle=\dfrac{4x^3\left(\frac{1}{2}-\frac{1}{1+\ln x} \right)}{1+x^4} + \dfrac{\log(1+x^4)-\log 2 }{x(1+\ln x)^2}$

$\large\sf\red{\underline{\underline{Therefore ,}}}$

$\rm\displaystyle\dfrac{d}{dx}\left(\int _1 ^{x^{2}} \dfrac{2t}{1+t^2}dt.\int ^{\ln x} _1 \dfrac{1}{(1+t)^2}\right) =\rm \dfrac{4x^3\left(\frac{1}{2}-\frac{1}{1+\ln x} \right)}{1+x^4} + \dfrac{\log(1+x^4)-\log 2 }{x(1+\ln x)^2}$