So I'm solving this problem which gives me: integral of function 1/(x^2+4x+13) dx
However the second step shows : 1/(x+2)^2 + 3^2
How did it go from the (x^2+4x+13) to (x+2)^2 + 3^2 ???
Any help is greatly appreciated!
Calculus Question?
2016-03-02 3:05 pm
回答 (1)
2016-03-02 3:13 pm
x² + 4x + 13
= x² + 4x + 4 + 9
= (x² + 4x + 4) + 9
= [x² + 2(2)x + (2)²] + 9
= (x + 2)² + 3²
= x² + 4x + 4 + 9
= (x² + 4x + 4) + 9
= [x² + 2(2)x + (2)²] + 9
= (x + 2)² + 3²
收錄日期: 2021-04-18 14:36:40
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