Question

Find the integral of

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\int\dfrac{\left\{\ln(x)\right\}^2+1}{x}\,dx}$

$\mapsto\,\bold{Put\,\,\,ln(x)=t}$

$\mapsto\,\bold{\dfrac{dx}{x}=dt}$

So, our integral becomes

$\displaystyle\rm{\int\left({t}^{2}+1\right)dt}$

$\displaystyle\rm{=\int{t}^{2}\,dt+\int\,dt}$

$\displaystyle\rm{=\dfrac{{t}^{2+1}}{2+1}+t+C}$

$\displaystyle\rm{=\dfrac{{t}^{3}}{3}+t+C}$

$\displaystyle\rm{=\dfrac{\left\{\ln(x)\right\}^{3}}{3}+\ln(x)+C}$