I have this function :
$$f(x) = \int_1^x(t+1)^3\ln(t+3)\,dt$$
I need to justify why this function is strictly increasing for $x \geq 1$.
Can somebody please help me find the derivative in order to show that it is increasing?
I know that a function is increasing when the first derivative is greater than $0$.
$\textbf{Hint: }$ Use the Leibnitz Integral Rule to differentiate both sides of the equation $$f(x) = \int_1^x(t+1)^3\ln(t+3)\,dt$$ Hence, $$f'(x)=(x+1)^3\ln(x+3)$$ Also note that $\ln(x+3)>0$ and $(x+1)^3>0$ for $x\ge 1$. $$$$What does $\ln(x+3)>0$ and $(x+1)^3>0$ for $x\ge1$ imply about the sign of $f'(x)=(x+1)^3\ln(x+3)$ for $x\ge1$?
Well, it implies that for $x\ge 1$, $f'(x)>0$. Hence $f(x)$ is $\textbf{ strictly increasing}$ for $x\ge1$. $$$$ $\textbf{Edit: }$ $$$$How do we know that $\ln(x+3)>0$ for $x\ge 1$? $$$$Put $y=x+3$, hence $\ln(x+3)=\ln(y)$. $$$$ Now, $\ln(y)$ is an increasing function in its domain ie the value of $\ln(y)$ is increasing for successive values of $y$.
Also clearly $\ln(y)>0$ for $y>1$ $$\Rightarrow \ln(x+3)>0\text{ for } x+3>1$$
Thus, $\ln(x+3)>0$ for $x>-2$. $$$$Hence for any value of $x$ which is greater than $-2$, $\ln(x+3)>0$. $$$$Similarly one can prove that $(x+1)^3>0$ for $x>0$ (and consequently for $x\ge1$).