como fazer integral indefinida integral x^2-3x-7/(2x+3) (x+1)^2 dx

Para resolver a integral dada, usaremos o método de Frações Parciais:

\(\begin{align} & \int_{{}}^{{}}{f}x=\int_{{}}^{{}}{\frac{{{x}^{2}}-3x-7}{(2x+3){{(x+1)}^{2}}}} \\ & \frac{{{x}^{2}}-3x-7}{(2x+3){{(x+1)}^{2}}}=\frac{A}{2x+3}+\frac{B}{x+1}+\frac{C}{{{(x+1)}^{2}}} \\ & {{x}^{2}}-3x-7=A{{(x+1)}^{2}}+B(2x+3)(x+1)+C(2x+3) \\ & {{x}^{2}}-3x-7=(A+2b){{x}^{2}}+(2A+5B+2C)x+A+3B+3C \\ \end{align} \)

Resolveremos o sistema obtido :

\(\begin{align} & A+2B=1 \\ & 2A+5B+2C=-3 \\ & A+3B+3C=-7 \\ & A=-1 \\ & B=1 \\ & C=-3 \\ \end{align} \)

Com as variáveis encontradas, calcularemos a integral :

\(\begin{align} & \int_{{}}^{{}}{f}x=\int_{{}}^{{}}{\frac{{{x}^{2}}-3x-7}{(2x+3){{(x+1)}^{2}}}} \\ & \int_{{}}^{{}}{\frac{{{x}^{2}}-3x-7}{(2x+3){{(x+1)}^{2}}}}=\int_{{}}^{{}}{\frac{-1}{2x+3}+\frac{1}{x+1}-\frac{3}{{{(x+1)}^{2}}}} \\ & \int_{{}}^{{}}{\frac{-1}{2x+3}+\frac{1}{x+1}-\frac{3}{{{(x+1)}^{2}}}}=\frac{-\ln (2x+3)}{2}+\ln (x+1)+\frac{3}{x+1} \\ & \int_{{}}^{{}}{\frac{-1}{2x+3}+\frac{1}{x+1}-\frac{3}{{{(x+1)}^{2}}}}=\frac{-\ln (2x+3)}{2}+\ln (x+1)+\frac{3}{x+1}+C \\ \end{align} \)

Portanto, a integral dada será:

\(\boxed{\int_{}^{} {\frac{{ - 1}}{{2x + 3}} + \frac{1}{{x + 1}} - \frac{3}{{{{(x + 1)}^2}}}} = \frac{{ - \ln (2x + 3)}}{2} + \ln (x + 1) + \frac{3}{{x + 1}} + C}\)