Question

The value of $\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^x\frac{t\ln (1+t)}{t^4+4}dt$ is

• A. $0$
• B. $\frac{1}{12}$
  
• C. $\frac{1}{24}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{1}{12}$

$\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^x\frac{t\ln (1+t)}{t^4+4}dt[\frac{0}{0}-form]$
Applying L' Hospital's rule, we get
$\underset{x\to 0}{lim}\frac{\frac{x\ln (1+x)}{x^4+4}}{3x^2}=\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}.\frac{1}{3(x^4+4)}=1.\frac{1}{12}=\frac{1}{12}$