Question

根号x分之1的不定积分为多少？

Answer 1

你好！直接套幂函数积分公式∫(1/√x)dx=∫x^(-1/2)dx=[1/(1+(-1/2))]x^(1+(-1/2))+c=2√x+c。经济数学团队帮你解答，请及时采纳。谢谢！

Answer 2

∫
1/√x
dx
=
∫
x^(-1/2)
dx
=
x^(-1/2+1)
/
(-1/2+1)
+
C
=
x^(1/2)
/
(1/2)
+
C
=
2√x
+
C