what's the integral of x/1+x^4 dx ?

Answer 1

Let u = x². Then du = 2x dx and we have:

∫x/(x⁴ + 1) dx = 0.5 ∫1/(u² + 1) du

Now we use the trig substitution u = tan z. Then, du = sec² z dz and we have:

0.5 ∫1/(u² + 1) du = 0.5 ∫ (sec² z)/(tan²z + 1)dz = 0.5 ∫ (sec² z)/(sec²z)dz = 0.5 ∫ dz = 0.5z + C

Now, we have to substitute backwards back to x. From the trig substitution, we know z = arctan(u). Therefore:

0.5z + C = 0.5arctan(u) + C = 0.5arctan(x²) + C

Therefore, our final answer is 0.5arctan(x²) + C