Question

pelo método da integral por substituição, me ajude!

a) integral de 3x^2 e^x3 dx
b) integral de x^2 e^x3 dx
c) x(x^2 + 1)^5 dx

Answer 1

a) 
$\displaystyle \int\,3x^2e^{x^3}\,dx\longrightarrow\ x^3=u\longrightarrow \int3x^2e^u\,dx\\\\ \frac{du}{dx}=3x^2\implies 3x^2dx=du\longrightarrow \int e^u\,du=e^u+C=\boxed{e^{x^3}+C}$

b)
$\displaystyle \int\,x^2e^{x^3}\,dx\longrightarrow u=x^3\longrightarrow \int\,x^2e^{u}\,dx\implies\\\\ u=x^3\implies \frac{du}{dx}=3x^2\implies du=3x^2dx\implies \underline{\frac{du}{3}=x^2dx}\\\\ \int\,x^2e^{u}\,dx=\int\,e^u\frac{du}{3}=\frac{1}{3}\int\,e^udu=\frac{1}{3}e^{u}+C=\boxed{\frac{1}{3}e^{x^3}+C}$

c)
$\displaystyle \int\,x(x^2+1)^{5}\,dx\longrightarrow u=x^2+1\implies \frac{du}{dx}=2x\implies xdx=\frac{du}{2}\\\\ \int\,x(u)^5\,dx=\int\,u^5\frac{du}{2}=\frac{1}{2}\int\,u^5\,du=\frac{1}{2}\cdot\frac{u^6}{6}+C=\boxed{\frac{(x^2+1)^6}{12}+C}$