Evaluating an Improper integral with square roots in the denominator and numerator

  • Thread starter heisgirl20
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I've tried to used integration by parts and u substitution and I've also tried just multiplying the fraction by the denominator (6-x)^(1/2) but I am still confused at how to approach this.
 
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  • #2
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Multiply and divide by [tex]\sqrt{6+x}[/tex].
 

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