一个简单的不定积分,如图

∫ x/(2 + x^4) dx
= 1/2 ∫ 1/(2 + x^4) d(x²)
= 1/2 ∫ 1/[(√2)² + (x²)²] d(x²)
= 1/2 * (1/√2) arctan(x²/√2) + C
= 1/(2√2) arctan(x²/√2) + C
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详细的解法：
∫ x/(2 + x^4) dx
令u = x²,du = 2x dx => dx = du/(2x)
= ∫ x/(2 + u²) * du/(2x)
= 1/2 ∫ du/(2 + u²)
令u = √2 tanz,du = √2 sec²z dz
= 1/2 ∫ (√2 sec²z)/(2 + 2tan²z) dz
= 1/√2 ∫ sec²z/(2sec²z) dz
= 1/(2√2) ∫ dz
= 1/(2√2) z + C
= 1/(2√2) arctan(u/√2) + C
= 1/(2√2) arctan(x²/√2) + C

原式=1/2∫d(x^2)/[2+(x^2)^2]=√2arctan(√2/2*x^2)/4+C