How do I evaluate $int \ ln(2x+1) \ dx$ ?

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Answer 1 · 2018-04-10T21:15:10

$int \ ln(2x+1) \ dx = 1/2(2x+1)ln(2x+1)-x+C$

Note that:

$d/dx((2x+1) ln(2x+1)) = 2ln(2x+1)+(2(2x+1))/(2x+1)$

$color(white)(d/dx((2x+1) ln(2x+1))) = 2ln(2x+1)+2$

So:

$d/dx(1/2(2x+1)ln(2x+1)-x) = ln(2x+1)$

So:

$int \ ln(2x+1) \ dx = 1/2(2x+1)ln(2x+1)-x+C$

How do I evaluate int \ ln(2x+1) \ dxint \ ln(2x+1) \ dx?