5 HHTerpaJI int (x^3dx)/(1+x^8)

To solve the integral \(\int \frac{x^3 \, dx}{1 + x^8}\), we can use a substitution method. Let's set \(u = x^4\). Then, \(du = 4x^3 \, dx\), which implies \(x^3 \, dx = \frac{1}{4} \, du\).<br /><br />Now, rewrite the integral in terms of \(u\):<br /><br />\[<br />\int \frac{x^3 \, dx}{1 + x^8} = \int \frac{1}{4} \cdot \frac{1}{1 + u^2} \, du<br />\]<br /><br />This simplifies to:<br /><br />\[<br />\frac{1}{4} \int \frac{1}{1 + u^2} \, du<br />\]<br /><br />The integral \(\int \frac{1}{1 + u^2} \, du\) is a standard form and equals \(\arctan(u)\). Therefore, we have:<br /><br />\[<br />\frac{1}{4} \int \frac{1}{1 + u} \, du = \frac{1}{4} \arctan(u) + C<br />\]<br /><br />Finally, substitute back \(u = x^4\):<br /><br />\[<br />\int \frac{x^3 \, dx}{1 + x^8} = \frac{1}{4} \arctan(x^4) + C<br />\]<br /><br />So, the solution to the integral is:<br /><br />\[<br />\boxed{\frac{1}{4} \arctan(x^4) + C}<br />\]