Question

usando um método apropriado, resolva a integral :

$\displaystyle\int \sf{ \dfrac{ x - (\arctan(x))^4 }{1 + x^2 } dx }$

Att: Joaquim Murrima ​

Answer 1

$\displaystyle\sf{\int\dfrac{x-(arctg(x))^4}{1+x^2}~dx=\int\dfrac{x}{1+x^2}~dx-\dfrac{(arctg(x))^4}{1+x^2}~dx}$

$\displaystyle\sf{\int\dfrac{x}{1+x^2}~dx=\dfrac{1}{2}\int\dfrac{2x}{1+x^2}~dx}\\\sf{fac_{\!\!,}a~t=1+x^2\implies~dt=2x~dx}\\\displaystyle\sf{\dfrac{1}{2}\int\dfrac{2x}{1+x^2}~dx=\dfrac{1}{2}\ell n\left|t\right|+k}\\\displaystyle\sf{\int\dfrac{x}{1+x^2}~dx=\dfrac{1}{2}\ell n\left|1+x^2\right|+k}$

$\displaystyle\rm{\int\dfrac{(arctg(x))^4}{1+x^2}~dx}\\\rm{fac_{\!\!,}a~w=arctg(x)\implies~dw=\dfrac{1}{1+x^2}~dx}\\\displaystyle\rm{\int\dfrac{(arctg(x))^4}{1+x^2}~dx=\int w^4~dw=\dfrac{1}{5}w^5+k}\\\displaystyle\rm{\int\dfrac{(arctg(x))^4}{1+x^2}~dx=\dfrac{1}{5}~arctg^5(x)+k}$

$\tt{portanto}\\\displaystyle\sf{\int\dfrac{x-(arctg(x))^4}{1+x^2}~dx=\dfrac{1}{2}\ell n\left|1+x^2\right|-\dfrac{1}{5}~arctg^5(x)+k}$