# How do you use the second fundamental theorem of Calculus to find the derivative of given int dx/lnx from [x^2, x^3]?

Sep 2, 2016

$= \frac{x \left(x - 1\right)}{\ln x}$

#### Explanation:

you have $\frac{d}{\mathrm{dx}} {\int}_{{x}^{2}}^{{x}^{3}} \frac{\mathrm{dx}}{\ln} x$ which is pretty ugly as the variable x is overloaded, so we can first re-write as

$\frac{d}{\mathrm{dx}} {\int}_{{x}^{2}}^{{x}^{3}} \frac{1}{\ln} t \mathrm{dt} q \quad \square$

in its simplest form, the Second FTC states that

$\frac{d}{\mathrm{dx}} {\int}_{a}^{x} f \left(t\right) \mathrm{dt} = f \left(x\right)$

BUT, in this case, we need to apply the Chain Rule as the interval is a function of x, so:

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} {\int}_{a}^{u \left(x\right)} f \left(t\right) \mathrm{dt} = f \left(u \left(x\right)\right) u ' \left(x\right)} q \quad \triangle$

Finally, to match $\square$ to $\triangle$ we do the following

${\int}_{{x}^{2}}^{{x}^{3}} \frac{1}{\ln} t \mathrm{dt}$

$= {\int}_{{x}^{2}}^{a} \frac{1}{\ln} t \mathrm{dt} + {\int}_{a}^{{x}^{3}} \frac{1}{\ln} t \mathrm{dt}$

$= - {\int}_{a}^{{x}^{2}} \frac{1}{\ln} t \mathrm{dt} + {\int}_{a}^{{x}^{3}} \frac{1}{\ln} t \mathrm{dt}$

$= {\int}_{a}^{{x}^{3}} \frac{1}{\ln} t \mathrm{dt} - {\int}_{a}^{{x}^{2}} \frac{1}{\ln} t \mathrm{dt}$

and now applying $\triangle$!!

$= \frac{1}{\ln} \left({x}^{3}\right) \frac{d}{\mathrm{dx}} \left({x}^{3}\right) - \frac{1}{\ln} \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$= \frac{3 {x}^{2}}{\ln} \left({x}^{3}\right) - \frac{2 x}{\ln} \left({x}^{2}\right)$

$= \frac{x \left(x - 1\right)}{\ln x}$