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# 1 integration Q(What's Wrong?)

Q:Integrate ∫dx/√(2-5x)

A:

Put √(2-5x)=u

2-5x=u²

5x=2-u²

x=(1/5)(2-u²)

dx/du=(1/5)(-2u)

dx=(-2u/5)du

So,

∫dx/√(2-5x)

=∫(-2u/5)du/u

=∫(-2/5)du

=0

WHAT'S WRONG??

### 1 Answer

Rating

- 翻雷滾天 風卷殘雲Lv 79 years agoFavorite Answer
∫(-2/5) du is not zero, instead:

∫(-2/5) du = -2u/5 + C since -2/5 is a constant

= -(2/5) √(2 - 5x) + C

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