In the evaluation of ∫dx√−x2−x+2 using Euler's substitution, which of the following is correct?
A
As the leading coefficient of the quadratic a and the vertical translation have different signs, polynomial can be factorised using real roots, first Euler substitution is used.
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B
As the leading coefficient of the quadratic a>0, First Euler substitution is used.
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C
As the leading coefficient of the quadratic a and the vertical translation have different signs, polynomial can be factorised using real roots, Second Euler substitution is used.
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D
As the leading coefficient of the quadratic a>0, Second Euler substitution is used.
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Solution
The correct option is C As the leading coefficient of the quadratic a and the vertical translation have different signs, polynomial can be factorised using real roots, Second Euler substitution is used.
If c>0 we take √ax2+bx+c=xt±√c.
We solve for x similarly as above and find, x=±2t√c−ba−t2.
Again, either the positive or the negative sign can be chosen.