Proving that $g(x)=\int_{0}^{\infty}f(t,x)dt$ is continuous

Question

I have the following function:
$$f(t,x)=\frac{log(x+t)}{(x+t)^2}$$ defined on $]0,\infty[\times ]0,\infty[$. I know that $t\rightarrow f(t,x)$ is an improper riemann integrable on $]0,\infty[$.
I want to prove that $g:]0,\infty[\rightarrow \mathbb{R}$ defined as: $$g(x)=\int_{0}^{\infty}f(t,x)dt$$ Is a continuous function. I tries doing this using the theorem that if there exists a improper riemann integrable $h:]0,\infty[\rightarrow \mathbb{R}$ with

$$|f(t,x)|\leq h(x)\text{ for every }(t,x)\in ]0,\infty[\times ]0,\infty[$$

then $g(t)$ must be continuous.
The problem is that I can't seem to find a function $h(x)$ such that $$|\frac{log(x+t)}{(x+t)^2}|\leq h(x)$$
Can anyone help me solve this?

Answer 1

What if we go directly by computation:

\begin{align*} \int_{0}^{M}\frac{\log(x+t)}{(x+t)^{2}}dt&=\frac{1}{(x+M)^{3}}-\frac{1}{x^{3}}-\int_{0}^{M}\frac{1}{x+t}\cdot-\frac{1}{x+t}dt\\ &=\frac{1}{(x+M)^{3}}-\frac{1}{x^{3}}-\frac{1}{x+M}+\frac{1}{x}, \end{align*} taking limit as $M\rightarrow\infty$, we have $g(x)=x^{-1}-x^{-3}$.