How do you use second fundamental theorem of calculus to find the derivative of given $\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt$ from [3, x]?

Hint: Now we are given with the integral $\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt$ . To solve this integral we will first use the method of substitution. To do so we will substitute ln(t) = u. Now we will get a simplified integral which can be solved by using the standard integral $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C$ . Now we know that according to second fundamental theorem of calculus we have if $F'\left( x \right)=f\left( x \right)$ then $\int\limits_{a}^{b}{f\left( x \right)}=F\left( b \right)-F\left( a \right)$ . Hence using this we will get the integral $\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt$ from [3, x].

Complete step-by-step solution:
Now first let us solve the given integration $\int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt$
To solve the integration we will use a method of substitution.
Let us substitute ln(t) = u.
Now differentiating the above equation on both sides we get,
$\Rightarrow \dfrac{1}{t}dt=du$
Now we will re-substitute the values of ln(t) and $\dfrac{1}{t}dt$ in the given integral. Hence, we get
$\Rightarrow \int{\left[ \dfrac{{{\ln }^{2}}\left( t \right)}{t} \right]}dt=\int{{{u}^{2}}du}$
Now we know the standard integral $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C$ Hence using this we get,
$\int{{{u}^{2}}}=\dfrac{{{u}^{3}}}{3}+C$
Now re-substituting the value of u we get,
\[\Rightarrow \int{\left[ \dfrac{\ln {{\left( t \right)}^{2}}}{t} \right]dt}=\dfrac{\ln {{\left( t \right)}^{3}}}{3}+C..............\left( 1 \right)\]
Now by second fundamental theorem of calculus we know that if we have $F'\left( x \right)=f\left( x \right)$ then $\int\limits_{a}^{b}{f\left( x \right)}=F\left( b \right)-F\left( a \right)$ . Hence using this theorem and equation (1) we get,
$\begin{align}
  & \Rightarrow \int_{3}^{x}{\dfrac{{{\ln }^{2}}\left( t \right)}{t}}dt=\dfrac{{{\ln }^{3}}\left( x \right)}{3}-\dfrac{{{\ln }^{3}}\left( 3 \right)}{3} \\
 & \Rightarrow \int_{3}^{x}{\dfrac{{{\ln }^{2}}\left( t \right)}{t}}dt=\dfrac{{{\ln }^{3}}\left( x \right)-{{\ln }^{3}}3}{3} \\
\end{align}$
Hence by second fundamental theorem of calculus we get $\int_{3}^{x}{\dfrac{{{\ln }^{2}}\left( t \right)}{t}}dt=\dfrac{{{\ln }^{3}}\left( x \right)-{{\ln }^{3}}3}{3}$


Note: Now keep a note that if we have $F'\left( x \right)=f\left( x \right)$ then $\int\limits_{a}^{b}{f\left( x \right)}=F\left( b \right)-F\left( a \right)$ and not $\int_{a}^{b}{F\left( x \right)}=f\left( b \right)-f\left( a \right)$ . Also while using the method of substitution always remember to re-substitute the expression by the original variable. Also while changing variables do not forget to change the derivative of the respective variable. For example if we substitute $\ln \left( t \right)=u$ then $du=\dfrac{1}{t}dt$ Hence do not just replace the differential dt by du.