Question

Me podrían ayudar a resolver estas integrales, por favoooor:

Answer 1

$</p><p>\int{\frac{dx}{2\sqrt{x}}}\\\texttt{Reexpresando:}\\\\\int{\frac{dx}{2\sqrt{x}}}=\frac{1}{2}\int{\frac{dx}{\sqrt{x}}}=\frac{1}{2}\int{\frac{dx}{x^{1/2}}}=\frac{1}{2}\int{x^{-1/2}}\,dx\\\texttt{Aplicamos la regla de la potencia}\\\texttt{para integrales}\\\\\frac{1}{2}\int{x^{-1/2}}\,dx=\frac{1}{2}\frac{x^{-1/2}+1}{-1/2+1}=\frac{1}{2}\frac{x^{1/2}}{1/2}=\frac{1}{2}*2x^{1/2}=x^{1/2}+C=\sqrt{x}+C\\\\\int{(x+b)^{3}}\,dx\\\texttt{Haciendo u=x+b,se obtiene:}\\u=x+b\,\,du=dx\\\int{(x+b)^{3}}\,dx=\int{u^{3}}\,du\\\texttt{Aplicando nuevamente la regla de las potencias:}\\\int{u^{3}}\,du=\frac{u^{3}+1}{3+1}=\frac{u^{4}}{4}+C\\\texttt{Deshaciendo el cambio:}\\\frac{u^{4}}{4}+C=\frac{(x+b)^{4}}{4}+C$

$\int{\frac{x}{x^{2}-1}}\,dx\\\texttt{Haciendo el cambio}\,\,u=x^{2}-1\,\,\texttt{nos queda:}\\u=x^{2}-1\,\,,du=2x\\frac{du}{2}=xdx\\\int{\frac{x}{x^{2}-1}}\,dx=\int{\frac{\frac{du}{2}}{u}=\int{\frac{du}{2u}}\\\texttt{Esta es una integral inmediata,por lo tanto:}\\\\\int{\frac{du}{2u}}=\frac{1}{2}\int{\frac{du}{u}}=\frac{1}{2}Ln(u)+C\\\texttt{Deshaciendo el cambio:}\\\\\frac{1}{2}Ln(u)+C=\frac{1}{2}Ln(x^{2}-1)+C$

$\int{e^{x^{2}}*\frac{x}{3}}\\\texttt{Haciendo}\,\,u=x^{2}\\\du=2x\,dx\\\frac{du}{2}=x\,dx\\\\\int{e^{x^{2}}*\frac{x}{3}}\,dx=\frac{1}{3}\int{e^{x^{2}}*x}\,dx=\frac{1}{3}\int{e^{u}*\frac{du}{2}}=\frac{1}{3}*\frac{1}{2}\int{e^{u}\,du}=\frac{1}{6}e^{u}+C\\\texttt{Deshaciendo el cambio:}\\\\\frac{1}{6}e^{u}+C=\frac{1}{6}e^{x^{2}}+C$

Saludos