Matemática, perguntado por marcelo7197, 8 meses atrás

usando um método apropriado, resolva a integral :

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

Att: Joaquim Murrima ​

Soluções para a tarefa

Respondido por CyberKirito
7

\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}


marcelo7197: perfeito mano...
CyberKirito: vlw mano :)
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