Question

41-46 use the comparison theorem to determine whether the given integral is convergent or divergent using comparison theorem.

\(\int_1^\infty {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx\)

Short answer

The integral function \(\int_1^\infty {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx\) is divergent.

Step-by-step solution

Solve the value of the integral\(\int_1^\infty  {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx\)

The integral function as follows:

\(\begin{aligned}{l}\int_1^\infty {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx &= \int_1^2 {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx + \int_2^\infty {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx\\\int_1^\infty {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx &= {I_1} + {I_2}\end{aligned}\)

Consider \({I_2}\) and let \(f(x) = \frac{{x + 1}}{{\sqrt {{x^4} - x} }}\).

Let \(g(x) = \frac{1}{x}\). It is known that \(\frac{{x + 1}}{{\sqrt {{x^4} - x} }} > \frac{1}{x}\).

Find whether \(\int_2^\infty g (x)dx\) is convergent or divergent.

\(\int_2^\infty g (x)dx = \int_2^\infty {\frac{1}{x}} dx\)

Use theorem \(\int_2^\infty {\frac{1}{{{x^p}}}} dx\) is convergent if \(p > 1\). and divergent if \(p \le 1\).

Thus, the integral function \(\int_2^\infty g (x)dx\) is divergent.

Solve the integral by using comparison theorem

Comparison Theorem:

Assume that\(f\)and\(g\)are continuous functions with\(f(x) \ge g(x) \ge 0\)for\(x \ge a\).

(a) If\(\int_a^\infty f (x)dx\)is convergent, then\(\int_a^\infty f (x)dx\)is convergent.

(b) If\(\int_a^\infty f (x)dx\)is divergent, then\(\int_a^\infty f (x)dx\)is divergent

Refer to the Comparison Theorem,

The integral function \(f(x)\) and \(g(x)\) are continuous functions and \(f(x) \ge g(x)\).

The integral function \(\int_2^\infty g (x)dx\) is divergent, therefore \(\int_2^\infty f (x)dx\) is divergent.

Since \({I_2}\) is divergent, \({I_1} + {I_2}\) also diverges.

Thus, the integral function\(\int_1^\infty {\frac{{x + 1}}{{\sqrt {{x^4} - x} }}} dx\)is divergent.