Question

Aplique o método da substituição para resolver as seguintes integrais:

a) $\int\limits^\frac{1}{2} _\frac{1}{3} {\frac{e^\frac{1}{x} }{x^2} } \, dx$


b) $\int\limits^2_1 {\frac{x^2}{(x^3+1)^2} } \, dx$

Answer 1

a)

$\huge\mathtt{\int\limits_{\frac{1}{3}}^{\frac{1}{2}}\dfrac{{e}^{\frac{1}{x}}}{{x}^{2}}dx}=\mathtt{-\int\limits_{\frac{1}{3}}^{\frac{1}{2}}-\dfrac{{e}^{\frac{1}{x}}}{{x}^{2}}dx}$

$\mathtt{u=\dfrac{1}{x}\to\,du=-\dfrac{1}{{x}^{2}}dx}\\\mathtt{se\,x=\frac{1}{2}\to\,u=2}\\\mathtt{se\,x=\frac{1}{3}\to\,u=3}$

$\mathtt{-\int\limits_{\frac{1}{3}}^{\frac{1}{2}}-\dfrac{{e}^{\frac{1}{x}}}{{x}^{2}}dx}=\mathtt{-\int\limits_{3}^{2}}({e}^{u})du}=\mathtt{\int\limits_{2}^{3}({e}^{u})du}$

$\mathtt{\int\limits_{2}^{3}{e}^{u}du={e}^{u}\big|_{2}^{3}={e}^{3}-{e}^{2}}$

b)

$\mathtt{\int\limits_{1}^{2}\dfrac{{x}^{2}}{{({x}^{3}+1)}^{2}}dx}\\=\mathtt{\dfrac{1}{3}\int\limits_{1}^{2}\dfrac{3{x}^{2}}{{({x}^{3}+1)}}dx}$

$\mathtt{u={x}^{3}+1\to\,du=3{x}^{2}dx}\\\mathtt{se\,x=1\to\,u=2}\\\mathtt{se\,x=2\to\,u=9}$

$\mathtt{\dfrac{1}{3}\int\limits_{1}^{2}\dfrac{3{x}^{2}}{{({x}^{3}+1)}}dx}}=\mathtt{\dfrac{1}{3}\int\limits_{2}^{9}\dfrac{du}{u}=\dfrac{1}{3}ln|u|\big|_{2}^{9}}$

$\mathtt{\dfrac{1}{3}ln|9|-\dfrac{1}{3}ln|3|=ln|{\dfrac{9}{3}}|^{\frac{1}{3}}=ln\sqrt[3]{3}}$