Asked by Anonymous
Integration with U-Substitution
Integral of [x/sqrt(x+1)] dx
I'm getting confused with how to deal with the x when I find the derivative of (x+1) because they don't match.
Integral of [x/sqrt(x+1)] dx
I'm getting confused with how to deal with the x when I find the derivative of (x+1) because they don't match.
Answers
Answered by
Steve
let
u^2 = x+1
x = u^2-1
2u du = dx
and we have
∫ (u^2-1)/u * 2u du
= 2∫u^2-1 du
and that's a cinch, right?
u^2 = x+1
x = u^2-1
2u du = dx
and we have
∫ (u^2-1)/u * 2u du
= 2∫u^2-1 du
and that's a cinch, right?
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