The value of \underset{x \rightarrow 0}{Lim} \textrm{ }\textrm{ } \frac{1}{x^{3}} \int_{0}^{x} \frac{t \textrm{ } ln \left(\right. 1 + t \left.\right)}{t^{4} + 4} \textrm{ } dt is

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

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

$\frac{1}{12}$

$\frac{1}{64}$

$\frac{1}{24}$

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$\underset{x \to 0}{Lim} \frac{\int_{0}^{x} \frac{t \ln (1 + t)}{t^{4} + 4} dt}{x^{3}} (\frac{0}{0}) form$

Applying L'Hospital's Rule once

$\underset{x \to 0}{Lim} \frac{\frac{xln (1 + x)}{(x^{4} + 4)}}{3 x^{2}} = \underset{x \to 0}{Lim} \frac{1}{3} \frac{\ln (1 + x)}{x} \frac{1}{4 + x^{4}} = \frac{1}{3} \cdot 1 \cdot \frac{1}{4 + 0} = {\frac{1}{12}}_{}$

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