Solución:
Desarrollamos la función a integrar
(xm+xn)2√x=x2m+2xm+n+x2nx12=x2m−12+2xm+n−12+x2n−12=x4m−12+2x2m+2n−12+x4n−12
Integramos
∫(xm+xn)2√xdx=∫(x4m−12+2x2m+2n−12+x4n−12)dx=x4m+124m+12+2x2m+2n+122m+2n+12+x4n+124n+12+c=2x4m+124m+1+4x2m+2n+122m+2n+1+2x4n+124n+1+c
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