Please view the images attached. To evaluate the integral |-(1+x)u = (1+x)", v'= In xb. u= (1+ x)", v' = In xc. u= In (1+ x), v'= (1+x)"d. u= In (1+ x), v' = (1+x)'In (1+x), v= (1+x)*-dx by Integration by Parts, the convenient choice isа.с.е.(1+ x) For computing the integral [x'ln (1+ x) dx , the most efficient method isa. Integration by Parts with u = In (1+x*) and v' = x'.substitution of u=1+x followed by Integration by Parts.twice Integration by Partsb.с.d. Integration by Parts with u'= In (1+x) and v= x.е.None of the methods covered so far is efficient in computing this integral.

Name: Please view the images attached. To evaluate the integral |-(1+x)u = (
Uploaded: 2024-03-20T07:07:37.000Z
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Description: Please view the images attached. To evaluate the integral |-(1+x)u = (1+x)", v'= In xb. u= (1+ x)", v' = In xc. u= In (1+ x), v'= (1+x)"d. u= In (1+ x), v' = (1+x)'In (1+x), v= (1+x)*-dx by Integration by Parts, the convenient choice isа.с.е.(1+ x) F

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Math

Calculus

To evaluate the integral -dx by Integration by Parts, the convenient choice is (1+ x)" a. u= (1+x)", v'= In x b. u= (1+x)', v'= In x c. u= In (1+ x), v'= (1+ x) d. u= In (1+ x), v'= (1+x)' In (1+x) с. . v' = (1+x)* е. (1+ x)