Does the following integral converge: $\int_6^{\infty}\frac{dx}{\sqrt{1+x^2}}$

Question

Does the following integral converge: $$\int_6^{\infty}\frac{dx}{\sqrt{1+x^2}}$$

I suppose we have to solve such problems by comparison test.

All the integrals I tried so far do not fit the test.

Can anyone show how to approach this.

Answer 1

Notice that

$$\frac{1}{\sqrt{1+x^2}}\sim_\infty\frac 1x$$ so the given integral is divergent.

Answer 2

You have, for $x>0$, $$ \sqrt{1+x^2}\leq \sqrt{1+2x+x^2}= \sqrt{(x+1)^2}=x+1 $$ thus $$ \int_6^\infty\frac{1}{\sqrt{1+x^2}}dx\geq \int_6^\infty\frac{1}{x+1}dx=\infty $$ and your integral is divergent.

Answer 3

If you evaluate the integral using the substitution $$x=tan(\theta)$$

the result is integration of $$sec(\theta)$$ with changed limits of integration and if evalueted completely it tends to infinity

Answer 4

Hint: Clearly, $x>6>1>0$, therefore $\underbrace{\sqrt{0+x^2}}_x<\sqrt{1+x^2}<\underbrace{\sqrt{x^2+x^2}}_{x\sqrt2}$.