Question

Integral indefinida
∫ $\frac{x-7 x^{3}+ 2x^{4} }{ x^{4} } dx$

Answer 1

$\int\ \frac{x-7x^3+2x^4}{x^4}\ dx\\\\ \int\ \frac{x(1-7x^2+2x^3}{x^4}\ dx\\\\ \int\ \frac{1-7x^2+2x^3}{x^3}\ dx$

$\int \frac{1-7x^2+2x^3}{x^3} = \int \frac{A+xB+x^2C}{x^3}\\\\ 1-7x^2+2x^3 = A+xB+x^2C\\\\ \boxed{x=0}\\\\ 1-0+0 = A+0B+0C\\\\ \boxed{A=1}$

$1-7x^2+2x^3 = A+xB+x^2C\\\\ \boxed{x=1}\\\\ 1-7+2=1+B+C\\\\ \boxed{B=-C-5}\\\\ 1-7x^2+2x^3 = A+xB+x^2C\\\\ \boxed{x=-1}\\\\ 1-7-2=1+C+5+C\\\\ \boxed{C=-7}\\\\ B=-(-7)-5\\\\ \boxed{B=2}$

Então, temos que:

$\int \frac{1-7x^2+2x^3}{x^3} = \int\ \frac{1}{x^3}\ dx+\int\ \frac{2}{x^2}\ dx-\int \frac{7}{x}\ dx\\\\ \boxed{\int =-\frac{1}{2x^2}-\frac{2}{x}-7\ ln|x|+K}$