Question
First make a substitution and then use integration by parts to evaluate. The integral of (x)(ln(x+3))dx What do you substitute first?
Answers
let y = x+3
then dy =dx
(y-3) ln y dy
= y ln y dy - 3 ln y dy
= y ln y dy - 3/y
let u = ln y then dv = y dy
so v = (1/2)y^2 and du = dy/y
so
(1/2)y^2 ln y -(1/2)ydy -3/y
(1/2)y^2 ln y - (1/4)y^2 - 3/y
but remember y = x + 3 :)
then dy =dx
(y-3) ln y dy
= y ln y dy - 3 ln y dy
= y ln y dy - 3/y
let u = ln y then dv = y dy
so v = (1/2)y^2 and du = dy/y
so
(1/2)y^2 ln y -(1/2)ydy -3/y
(1/2)y^2 ln y - (1/4)y^2 - 3/y
but remember y = x + 3 :)
screw dis
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