Question

If $\int \frac{\left(8x+\frac{8}{x}+9\right)}{{\left(2x^3+3x^2+4x\right)}^2}dx=g(x)+C$, then the value of $9(g(1)-g(-1))$ is equal to

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$\int \frac{\left(8x+\frac{8}{x}+9\right)x^2}{{\left(2x^4+3x^3+4x^2\right)}^2}dx=\int \frac{\left(8x^3+9x^2+8x\right)}{{\left(2x^4+3x^3+4x^2\right)}^2}dx$
Put $2x^4+3x^3+4x^2=t\Rightarrow dt=\left(8x^3+9x^2+8x\right)dx$
$\int \frac{dt}{t^2}=\frac{-1}{t}+C=\frac{-1}{2x^4+3x^3+4x^2}+C$
$g(1)=\frac{-1}{9},g(-1)=\frac{-1}{3}$
$g(1)-g(-1)=\frac{-1}{9}+\frac{1}{3}=\frac{2}{9}$
$9(g(1)-g(-1))=2$