Question

Calcule a integral indefinida.


$\mathsf{\displaystyle\int\dfrac{x+sen(x)-cos(x)-1}{x+e^x+sen(x)}~dx}$


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Por favor responder de forma detalhada.

Answer 1

$I=\mathsf{\displaystyle\int\dfrac{x+sen(x)-cos(x)-1}{x+e^x+sen(x)}~dx}\\\\I=\mathsf{\displaystyle\int\dfrac{(x+sen(x)+e^x)-e^x-cos(x)-1}{x+e^x+sen(x)}~dx}\\\\I=\mathsf{\displaystyle\int\dfrac{-e^x-cos(x)-1}{x+e^x+sen(x)}+1~dx}\\\\I= - \mathsf{\displaystyle\int\dfrac{e^x+cos(x)+1}{x+e^x+sen(x)}~dx+\int dx}\ \\\\I=\mathsf{-ln|x+e^x+sen(x)|+x+C}$

Answer 2

$\displaystyle\sf{\int\dfrac{x+sen(x)-cos(x)-1}{x+e^x+sen(x)}~dx}$

Somando e subtraindo $\sf{e^x}$ no numerador temos:

$\displaystyle\sf{\int\dfrac{\left[x+sen(x)+e^x\right]-e^x-cos(x)-1}{x+e^x+sen(x)}~dx}\\\sf{\displaystyle\int\dfrac{x+sen(x)+e^x}{x+sen(x)+e^x}~dx-\int\dfrac{e^x+cos(x)+1}{x+e^x+sen(x)}~dx}\\\displaystyle\sf{\int~dx-\int\dfrac{e^x+cos(x)+1}{x+e^x+sen(x)}~dx}$

faça:

$\sf{u=x+e^x+sen(x)\implies du=1+e^x+cos(x)dx}$

$\displaystyle\sf{\int\dfrac{e^x+cos(x)+1}{x+e^x+sen(x)}~dx=\int\dfrac{du}{u}=\ell n|u|+k}$

Portanto

$\boxed{\boxed{\displaystyle\sf{\int\dfrac{x+sen(x)-cos(x)+1}{x+e^x+sen(x)}~dx=x-\ell n|x+e^x+sen(x)|+k}}}$