Question

Calcule a integral indefinida usando as regras da substituição.

Answer 1

27. $\int{\dfrac{e^{x}}{1+e^{x}}\,dx}$

Substituição:

$u=1+e^{x}\;\;\Rightarrow\;\;du=e^{x}\,dx$

$\int{\dfrac{e^{x}}{1+e^{x}}\,dx}\\ \\ \\ =\int{\dfrac{du}{u}}\\ \\ \\ =\mathrm{\ell n\,}|u|+C\\ \\ \\ =\mathrm{\ell n\,}|1+e^{x}|+C$


37. $\int{\dfrac{1}{x\mathrm{\,\ell n\,}x}\,dx}$

Substituição:

$u=\mathrm{\ell n\,}x\;\;\Rightarrow\;\;du=\dfrac{dx}{x}\\ \\ \\ \int{\dfrac{1}{x\mathrm{\,\ell n\,}x}\,dx}\\ \\ \\ =\int{\dfrac{1}{\mathrm{\ell n\,}x}\cdot \dfrac{dx}{x}}\\ \\ \\ =\int{\dfrac{1}{u}\,du}\\ \\ \\ =\mathrm{\ell n\,}|u|+C\\ \\ \\ =\mathrm{\ell n\,}|\mathrm{\ell n\,}x|+C$


33. $\int{e^{2x}(e^{2x}+1)^{3}\,dx}$

Substituição:

$u=e^{2x}+1\;\;\Rightarrow\;\;du=2e^{2x}\,dx\;\;\Rightarrow\;\;e^{2x}\,dx=\dfrac{1}{2}\,du\\ \\ \\ \int{e^{2x}(e^{2x}+1)^{3}\,dx}\\ \\ \\ =\int{(e^{2x}+1)^{3}\,e^{2x}\,dx}\\ \\ \\ =\int{u^{3}\cdot \dfrac{1}{2}\,du}\\ \\ \\ =\dfrac{1}{2}\int{u^{3}\,du}\\ \\ \\ =\dfrac{1}{2}\cdot \dfrac{u^{3+1}}{3+1}+C\\ \\ \\ =\dfrac{1}{2}\cdot \dfrac{u^{4}}{4}+C\\ \\ \\ =\dfrac{u^{4}}{8}+C\\ \\ \\ =\dfrac{(e^{2x}+1)^{4}}{8}+C$


43. $\int{\dfrac{x+1}{\sqrt{x}-1}\,dx}$

Substituição:

$\sqrt{x}-1=u\;\;\Rightarrow\;\;\sqrt{x}=u+1\;\;\Rightarrow\;\;x=(u+1)^{2}\\ \\ \\ \;\;\Rightarrow\;\;dx=2\,(u+1)\,du\\ \\ \\ \int{\dfrac{x+1}{\sqrt{x}-1}\,dx}\\ \\ \\ =\int{\dfrac{(u+1)^{2}+1}{u}\cdot 2\,(u+1)\,du}\\ \\ \\ =2\int{\dfrac{(u^{2}+2u+1+1)\,(u+1)}{u}\,du}\\ \\ \\ =2\int{\dfrac{(u^{2}+2u+2)\,(u+1)}{u}\,du}\\ \\ \\ =2\int{\dfrac{u(u^{2}+2u+2)+1\,(u^{2}+2u+2)}{u}\,du}\\ \\ \\ =2\int{\dfrac{u^{3}+2u^{2}+2u+u^{2}+2u+2}{u}\,du}\\ \\ \\ =2\int{\dfrac{u^{3}+3u^{2}+4u+2}{u}\,du}\\ \\ \\ =2\int{\left(\dfrac{u^{3}+3u^{2}+4u}{u}+\dfrac{2}{u} \right )\,du}\\ \\ \\ =2\int{\left(\dfrac{\diagup\!\!\!\! u\,(u^{2}+3u+4)}{\diagup\!\!\!\! u}+\dfrac{2}{u} \right )\,du}$


$=2\int{(u^{2}+3u+4)\,du}+4\int{\dfrac{du}{u}}\\ \\ \\ =2\left(\dfrac{u^{2+1}}{2+1}+3\cdot \dfrac{u^{1+1}}{1+1}+4u \right )+4\mathrm{\,\ell n\,}|u|+C\\ \\ \\ =2\left(\dfrac{u^{3}}{3}+\dfrac{3u^{2}}{2}+4u \right )+4\mathrm{\,\ell n\,}|u|+C\\ \\ \\ =\dfrac{2u^{3}}{3}+3u^{2}+8u+4\mathrm{\,\ell n\,}|u|+C\\ \\ \\ =\dfrac{2(\sqrt{x}-1)^{3}}{3}+3(\sqrt{x}-1)^{2}+8(\sqrt{x}-1)+4\mathrm{\,\ell n\,}|\sqrt{x}-1|+C$


47. $\int{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\,dx}$

Substituição:

$1+\sqrt{x}=u\;\;\Rightarrow\;\;\sqrt{x}=u-1\;\;\Rightarrow\;\;x=(u-1)^{2}\\ \\ \Rightarrow\;\;dx=2\,(u-1)\,du\\ \\ \\ \int{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\,dx}\\ \\ \\ =\int{\dfrac{1-(u-1)}{u}\cdot2\,(u-1)\,du}\\ \\ \\ =2\int{\dfrac{(1-u+1)\,(u-1)}{u}\,du}\\ \\ \\ =2\int{\dfrac{(2-u)\,(u-1)}{u}\,du}\\ \\ \\ =2\int{\dfrac{u\,(2-u)-1\,(2-u)}{u}\,du}\\ \\ \\ =2\int{\dfrac{2u-u^{2}-2+u}{u}\,du}\\ \\ \\ =2\int{\dfrac{3u-u^{2}-2}{u}\,du}\\ \\ \\ =2\int{\dfrac{3u-u^{2}}{u}\,du}-4\int{\dfrac{du}{u}}\\ \\ \\ =2\int{\dfrac{\diagup\!\!\!\! u\,(3-u)}{\diagup\!\!\!\! u}\,du}-4\int{\dfrac{du}{u}}$


$=2\int{(3-u)\,du}-4\int{\dfrac{du}{u}}\\ \\ \\ =2\left(3u-\dfrac{u^{2}}{2} \right )-4\mathrm{\,\ell n\,}|u|+C\\ \\ \\ =6u-u^{2}-4\mathrm{\,\ell n\,}|u|+C\\ \\ \\ =6\,(1+\sqrt{x})-(1+\sqrt{x})^{2}-4\mathrm{\,\ell n\,}|1+\sqrt{x}|+C$