Integration - two ways two different answers?

trojan

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Feb 7, 2007
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I solved the following integration problem two different ways and got two different answers, but both ways look sound to me. Can someone see where I went wrong.

Integrate (x^2 + 3x + 7) / (sqrt x)

Using "u" subsitution, I'm able to get the answer that is in the back of the book and this chapter is also on "u" substituting.

Let u = x^(1/2) then dx = 2x^(1/2)du
Substituting into the integral above gives integral (u^4 +3u^2 +7) *(2u du) / (u). This solves out to 2/5 u^5 + 2 u^3 + 14 u +C. Then substitute back in x^(1/2) to get 2/5 x^(5/2) + 2 x^(3/2) +14 x^(1/2) + C

Okay now the other way I solved it, was the first way I tried. I did long divsion to rewrite the integral (x^2 + 3x + 7) dx / (sqrt x) = integral [x^4 + 3x^2 + 7 / x^(1/2)]dx. That integral evaluates out to 1/5 x^5 + x^3 + 14 x^(1/2) +C

So I have two different (but kind of similar) answers. The first one is right according to the book, but the second looks right to me as well. Any insight would be appreciated.
 
trojan said:
I did long divsion to rewrite the integral (x^2 + 3x + 7) dx / (sqrt x) = integral [x^4 + 3x^2 + 7 / x^(1/2)]dx. That integral evaluates out to 1/5 x^5 + x^3 + 14 x^(1/2) +C

Division before you integrate was wrong... Algebra mistake basically.

\(\displaystyle \L \frac{x^2 + 3x + 7}{x^{1/2}}
= x^{3/2} + 3x^{1/2} + 7x^{-1/2}\)

You get the same answer when you integrate this.
 
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