Question

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

• A. $0$
• B. $\frac{1}{12}$
• C. $\frac{1}{24}$
• D. $\frac{1}{64}$

Answer 1

The correct option is B

$\frac{1}{12}$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 0}{\lim }\frac{1}{x^3}{\int }_0^x\frac{t\ln \left(1+t\right)}{t^4+4}dt$

By Leibnitz's rule it can be written as,

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\frac{\frac{x\ln \left(1+x\right)}{x^4+4}}{3x^2} \\ \Rightarrow \frac{1}{3}\underset{x\to 0}{\lim }\frac{\ln \left(1+x\right)}{x(x^4+4)} \\ \Rightarrow \frac{1}{3}\underset{x\to 0}{\lim }\frac{\left({\displaystyle \frac{\ln \left(1+x\right)}{x}}\right)}{(x^4+4)}\left[\underset{x\to 0}{\lim }\frac{\ln \left(1+x\right)}{x}=1\right] \\ \Rightarrow \frac{1}{3}\left(\frac{1}{4}\right) \\ \Rightarrow \frac{1}{12} \end{gathered}$$

Therefore, the value of $\underset{x\to 0}{\lim }\frac{1}{x^3}{\int }_0^x\frac{t\ln \left(1+t\right)}{t^4+4}dt$ is $12$.

Hence, option $\left(B\right)$ is the correct option.