Question

Alguien que me ayude, por favor.


250 PUNTOS NETOS!!!


${\int {\dfrac{2x-5+\tan^{-1}(x)+ e^{arccot(x)}}{1+x^2}} \, dx }$

Answer 1

$\displaystyleI=\int {\dfrac{2x-5+\tan^{-1}(x)+ e^{arccot(x)}}{1+x^2}} \, dx \\ \\ \\I=\int \dfrac{2x}{1+x^2}~dx-\int\dfrac{5}{1+x^2}~dx+\int\dfrac{\arctan x}{1+x^2}~dx+\int\dfrac{e^{\arctan x}}{1+x^2}~dx\\ \\ \\I=\ln(1+x^2)-5\arctan{x}+\int \arctan{x} ~d(\arctan{x})+\int e^{\arctan{x}}~d(\arctan{x})\\ \\ \\\boxed{I=\ln(1+x^2)-5\arctan{x}+\dfrac{1}{2}\arctan^2{x}+e^{\arctan{x}}+C}$