10 from 10 Solve the definite integral and enter your answer in decimal format (use the a/b form for irrational fractions) int _(-1)^1((2x)/(3)+(3)/(x^4))dx square

To solve the definite integral \(\int_{-1}^{1} \left( \frac{2x}{3} + \frac{3}{x^4} \right) \, dx\), we need to evaluate each term separately.<br /><br />First, let's consider the integral of the first term:<br />\[<br />\int_{-1}^{1} \frac{2x}{3} \, dx<br />\]<br /><br />Since \(\frac{2x}{3}\) is an odd function (i.e., \(f(-x) = -f(x)\)), the integral of an odd function over a symmetric interval around zero is zero:<br />\[<br />\int_{-1}^{1} \frac{2x}{3} \, dx = 0<br />\]<br /><br />Next, let's consider the integral of the second term:<br />\[<br />\int_{-1}^{1} \frac{3}{x^4} \, dx<br />\]<br /><br />Similarly, \(\frac{3}{x^4}\) is also an odd function. Therefore, its integral over the interval \([-1, 1]\) is also zero:<br />\[<br />\int_{-1}^{1} \frac{3}{x^4} \, dx = 0<br />\]<br /><br />Since both integrals are zero, the value of the original definite integral is:<br />\[<br />\int_{-1}^{1} \left( \frac{2x}{3} + \frac{3}{x^4} \right) \, dx = 0 + 0 = 0<br />\]<br /><br />Thus, the answer is:<br />\[<br />\boxed{0}<br />\]