# Which of the following improper integrals are convergent?

Which of the following improper integrals are convergent?

a.$\quad\displaystyle \int_{1}^{\infty} \frac{dx}{\sqrt{x^3+ 2x + 2}}$

b. $\quad\displaystyle \int_{0}^{5}\frac{dx}{(x^2− 5x + 6)}$

c.$\quad\displaystyle \int_{0}^{5} \frac{dx}{\sqrt[\large 3]{7x + 2x^4}}$

how can I able to solve this.thanks for your help

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The first one is convergent. Because $$\lim_{x\to\infty}x^{\frac{3}{2}}\cdot\frac{1}{\sqrt{x^3+2x+2}}=1$$ and we know when the resulted limit is finite when the power of $x$ is greater than $1$ (I mean that $x^{\frac{3}{2}}$ ), the improper integral $\int_a^{\infty}$ converges.
About the second. If we divide it to 3 parts, we have: $$\int_0^5=\int_0^2+\int_2^3+\int_3^5$$ Consider the first part: $$\int_0^2\frac{dx}{x^2-5x+6}=\int_0^2\frac{dx}{(x-2)(x-3)}$$ You see that if we consider $(x-2)^1$ and multiply it to the integrand $\frac{1}{(x-2)(x-3)}$ and take the limit of the result when $x\to 2^-$, then: $$\lim_{x\to 2^-}(x-2)^1\frac{1}{(x-2)(x-3)}=-1\neq0$$ whenever you can find a power like $1$ (I mean in $(x-2)^1$) and so find the above limit non zero, the improper integral diverges. So the second one is divergent.
The 2nd integral diverges because the interval of integration runs through both poles at $x=2$ and $x=3$, and the behavior at either pole $x_k$ is $1/x_k$. The 3rd converges because a) the integrated is nonnegative throughout the integration interval, and b) the behavior at $x=0$ is $x^{-1/3}$? Which is integrable.