Evaluate the integral by integration. Round your answer to two decimal places.

Question

Evaluate the integral  by integration. Round your answer to two decimal places. ​

Quizplus · Accepted Answer

We can use integration by substitution as follows:
Let $u = x^2 + 1$, then $du = 2xdx$
Thus, the integral becomes:
$$\int \frac{x}{(x^2+1)^2}dx = \frac{1}{2}\int \frac{1}{u^2}du = -\frac{1}{2u} + C = -\frac{1}{2(x^2+1)} + C$$
Evaluating the definite integral from 0 to 2 gives:
$$\left[-\frac{1}{2(x^2+1)}\right]_{0}^{2} = \left(-\frac{1}{2(2^2+1)}\right) - \left(-\frac{1}{2(0^2+1)}\right) \approx 0.71$$
Therefore, the answer is choice C, 0.71.

Evaluate the Integral by Integration