Question

Integral indefinida x^2 (3+x)^-4

Answer 1

integral indefinida

$<strong>\boxed{\int f(x)dx=g(x)+k} </strong>$

$\int {x}^{2}{(3+x)}^{-4}dx=\int \frac{{x}^{2}}{{(3+x)}^{4}}dx$

faça

$u=x+3 \\ x=u-3 \\ du=dx$

$\int \frac{{x}^{2}}{{(3+x)}^{4}}dx=\int \frac{{(u-3)}^{2}}{{u}^{4}}du$

$\int \frac{{(u-3)}^{2}}{{u}^{4}}du\\ \int {u}^{-4}({(u-3)}^{2})du\\\int {u}^{-4} ({u}^{2}-6u+9)du$

$\int ({u}^{-2}-6{u}^{-3}+9{u}^{-4})du\\=-\frac{1}{u}-\frac{3}{{u}^{2}}-\frac{3}{{u}^{3}}+k$

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