Question

The value of $\int_{}^{}\frac{2x^{3} - 3x^{2} + 5x + 6}{x^{2} + 3x + 2}dx$ is

• A. $x^{2} + 3x + 4\ln|x^{2} + 3x + 2| + 12\ln\frac{x + 1}{x + 2} + c$
• B. $x + 3x^{2} + 4\ln|x + 1| - 12\ln(x + 2) + c$
• C. $\left( x ^ { 2 } + 3 x \right) + 8$ $\ln\frac{|x + 1}{(x + 2)^{2}} + c$
• D. None of these

Answer 1

Answer: (C)

\left( x ^ { 2 } + 3 x \right) + 8$$\ln\frac{|x + 1}{(x + 2)^{2}} + c

Answer 2

$\int_{}^{}\frac{2x^{3} - 3x^{2} + 5x + 6}{x^{2} + 3x + 2}dx$

= $\int_{}^{}{\left( (2x + 3) - \frac{8x}{x^{2} + 3x + 2} \right)dx}$

=$\int_{}^{}{(2x + 3) - 4\int_{}^{}\frac{(2x + 3)dx}{x^{2} + 3x + 2} + 12\int_{}^{}\frac{dx}{(x + 1)(x + 2)}} = (x^{2} + 3x) - 4\ln|x^{2} + 3x + 2| + 12\int_{}^{}{\frac{dx}{x + 1} - 12}\int_{}^{}\frac{dx}{x + 2}$= $(x^{2} + 3x) - 4\ln(x + 1) - 4\ln(x + 2) + 12\ln\frac{x + 1}{x + 2} + c$ = $x^{2} - 3x + 8\ln(x + 1) - 16\ln(x + 2) + c$

= $(x^{2} - 3x) + 8\ln\frac{|x + 1|}{(x + 2)^{2}} + c$.