Question:

$\int \frac{2x^2 + 3x^5}{1+x^6} dx = \frac{2}{3} \int \frac{3x^2 dx}{1+x^6} + \frac{3}{2} \int \frac{6x^5 dx}{1+x^6}$<br>In first integral, put $x^3 = t \Rightarrow 3x^2 dx = dt$<br>In second integral, put $1 + x^6 = u \Rightarrow 6x^5 dx = du$<br>$\therefore \int \frac{2x^2 + 3x^5}{1+x^6} dx = \frac{2}{3} \int \frac{dt}{1+t^2} + \frac{1}{2} \int \frac{du}{u}$<br>$= \frac{2}{3} \tan^{-1}(t) + \frac{1}{2} \ln |u| + C$<br>$= \frac{2}{3} \tan^{-1}(x^3) + \frac{1}{2} \ln |1 + x^6| + C$