Show divergence of $\int_{1}^{\infty} \frac{x+\sqrt{x}+7}{x^2+2x+1} dx$

Question

$$\int_{1}^{\infty} \frac{x+\sqrt{x}+7}{x^2+2x+1} dx$$

How can I show that the following integral diverges by using the direct comparison test? Thanks in advance!

Answer 1

Note,

$$\int_{1}^{\infty} \frac{x+\sqrt{x}+7}{x^2+2x+1} dx>\int_{1}^{\infty} \frac{x}{(x+1)^2} dx =\int_{1}^{\infty} \frac{1}{x+1}dx-\int_{1}^{\infty} \frac{1}{(x+1)^2} dx$$ $$=\int_{1}^{\infty} \frac{1}{x+1}dx-\frac12$$

which is divergent because $\int_{1}^{\infty} \frac{1}{x+1}dx$ is divergent.

Answer 2

For sure, the problem is at the upper bound.

Let $x=y^2$ to make $$\int_{1}^{\infty} \frac{x+\sqrt{x}+7}{x^2+2x+1} dx=\int_{1}^{\infty} \frac{2 y \left(y^2+y+7\right)}{\left(y^2+1\right)^2}\,dy$$

Use the long division for the integrand $$\frac{2 y \left(y^2+y+7\right)}{\left(y^2+1\right)^2}=\frac{2}{y}+\frac{2}{y^2}+\cdots$$