Proof that $\int_0^{\Lambda}\frac{x^{d-1}}{(1+x^2)^2}dx$ remains finite for $1<d<4$

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For a phase transition in Landau theory, I need to show that

$$I=\int_0^{\Lambda}\frac{x^{d-1}}{(1+x^2)^2}dx$$ remains finite for $1<d<4$, as $\Lambda \rightarrow \infty$.

Now I know, that

$$\lim_{x\rightarrow 0} \frac{x^{d-1}}{(1+x^2)^2} \sim x^{d-1}$$

which remains regular for $d>1$,

and

$$\lim_{x\rightarrow \infty} \frac{x^{d-1}}{(1+x^2)^2} \sim x^{d-5}$$

which remains regular for $d<4$. But how do I get to use these approximations? I would have to integrate first, to apply the upper/lower limits to the integrand?

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In fact $0<d<4$ is enough. Split the integral into integral from $0$ to $1$ and $1$ to $\infty$. The first part converges for all $d>0$ because the integrand is bounded by $x^{d-1}$. Now $\int_1^{\infty} \frac {x^{d-1}} {(1+x^{2})^{2}}\, dx \leq \int_1^{\infty} x^{d-1-4}\, dx <\infty$ if $d <4$.