Question

If $\int \frac{(2x+3)(x^2-3x+1)dx}{(x^4-7x^2+1)(1+\ln (1+3x+x^2))}=f\left(x\right)+C,$ where $C$ is constant and $f\left(0\right)=0,$ then

• A. $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=3$
• B. $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=\frac{1}{3}$
• C. $\underset{x\to 0}{lim}\frac{f\left(x^2\right)}{\ln \left(\cos x\right)}=-6$
• D. $\underset{x\to 0}{lim}\frac{f\left(x^2\right)}{\ln \left(\cos x\right)}=-\frac{1}{6}$

Answer 1

Answer: (A), (C)

$\underset{x\to 0}{lim}\frac{f\left(x^2\right)}{\ln \left(\cos x\right)}=-6$

$I=\int \frac{(2x+3)(x^2-3x+1)dx}{(x^4-7x^2+1)(1+\ln (1+3x+x^2))}$
$=\int \frac{(2x+3)(x^2-3x+1)dx}{(x^2-3x+1)(x^2+3x+1)(1+\ln (1+3x+x^2))}$
$=\int \frac{(2x+3)dx}{(x^2+3x+1)(1+\ln (1+3x+x^2\left)\right)}$

Putting $1+\ln (1+3x+x^2)=t$
$\Rightarrow \frac{(2x+3)dx}{1+3x+x^2}=dt$
$\therefore I=\int \frac{dt}{t}$
$=\ln |t|+C$
$=\ln (1+\ln (1+3x+x^2))+C$
$\therefore f\left(x\right)=\ln (1+\ln (1+3x+x^2\left)\right)$


$L_1=\underset{x\to 0}{lim}\frac{\ln (1+\ln (1+3x+x^2\left)\right)}{x}\left(\frac{0}{0}\right)\text{form}$
By L' Hospital rule,
$L_1=\underset{x\to 0}{lim}\frac{0+\frac{2x+3}{(1+3x+x^2)}}{1+\ln (1+3x+x^2)}=3$


$L_2=\underset{x\to 0}{lim}\frac{f\left(x^2\right)}{\ln \left(\cos x\right)}$
$=\underset{x\to 0}{lim}\frac{\ln (1+\ln (1+3x^2+x^4\left)\right)}{\ln \left(\cos x\right)}$
$=\underset{x\to 0}{lim}\frac{\ln (1+\ln (1+3x^2+x^4\left)\right)}{\ln (1+3x^2+x^4)}\times \frac{\ln (1+3x^2+x^4)}{\ln \left(\cos x\right)}$
$=\underset{x\to 0}{lim}\frac{\ln (1+3x^2+x^4)}{3x^2+x^4}\times \frac{3x^2+x^4}{\ln \left(\cos x\right)}$
$=\underset{x\to 0}{lim}\frac{3x^2+x^4}{\ln \left(\cos x\right)}\left(\frac{0}{0}\right)\text{form}$
By L' Hospital rule, we have
$L_2=-\underset{x\to 0}{lim}\frac{(6x+4x^3)\cos x}{x\times \frac{\sin x}{x}}=-6$