91Ó°ÊÓ

To make the integral a proper one, substitute \( x = \frac{1}{t} \). This means \( dx = -\frac{1}{t^2} dt \). Substitute these into the integral: \[\int_{1}^{\infty} \frac{x}{x^3+1} dx = \int_{0}^{1} \frac{\frac{1}{t}}{\left(\frac{1}{t}\right)^3 + 1} \left(-\frac{1}{t^2}\right) dt\] Simplify the integrand to:\[-\int_{0}^{1} \frac{1}{t(t^3 + 1)} dt\] This is now a proper integral.

Rewrite the integrand:\[-\int_{0}^{1} \frac{1}{t^4 + t} dt = -\int_{0}^{1} \frac{1}{t(t^3 + 1)} dt\] This is the simplified form we will work with using Simpson's Rule.

Simpson's Rule states that for an integral \( \int_a^b f(x) \, dx \) with \( n \) intervals:\[\int_a^b f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{i=1, 3, 5, \ldots}^{n-1} f(x_i) + 2 \sum_{i=2, 4, 6, \ldots}^{n-2} f(x_i) + f(x_n) \right]\]where \( h = \frac{b-a}{n} \). In this case, \( a = 0 \), \( b = 1 \), and \( n = 4 \), giving \( h = 0.25 \). We calculate the following:- \( x_0 = 0\), \( x_1 = 0.25 \), \( x_2 = 0.5 \), \( x_3 = 0.75 \), \( x_4 = 1 \).- Evaluate \( f(x) = -\frac{1}{x(x^3 + 1)} \) at each point.After calculating these values:- \( f(x_0) = -1 \),- \( f(x_1) = -0.615 \),- \( f(x_2) = -0.267 \),- \( f(x_3) = -0.108 \),- \( f(x_4) = -0.5 \).Applying Simpson's Rule:\[\int_0^1 -\frac{1}{x(x^3 + 1)} \, dx \approx \frac{0.25}{3} [-1 + 4(-0.615) + 2(-0.267) + 4(-0.108) + (-0.5)] \approx -0.733\]Thus, the approximate value using Simpson's Rule is \(-0.733\).

The integral of \( \frac{x}{x^3+1} \) from 1 to infinity is approximately -0.733, indicating that the area under the curve is finite and the improper integral converges.