integral of ln(2x+1)dx using integration by parts method?

Question

so far i've got....u=ln(2x+1), du=2dx/(2x+1), dv=dx, v=x

xln(2x+1)-integ 2xdx/(2x+1)

Answer 1

Good start. I'd say you've gone as far as you can with IBP.

To finish, it calls for a substitution. You've already used u, so let's say w=2x+1, dw=2dx, 2x=w-1. You are now trying to find

(1/2)Int (w-1)dw/w, which simplifies to (1/2)Int [1-1/w]dw = (1/2)[w-ln|w|] = x+1/2-(1/2)ln|2x+1|.

I get a final answer of (x+1/2)ln(2x+1) - x + C.

Answer 2

You've got it right so far. Apply the formula,

uv - Int(vdu)

The integral will be xdx/(2x+1). Simplify this by long division and the rest is cake.

Answer 3

whether the integrand is obviously the made of two words integration by skill of aspects isn't a treatment all. evaluate integrating x sqrt(x^2+a million). Integration by skill of aspects isn't the clever mindset; an trouble-free substitution is.