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

Answer 1

Scarlett Allison

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

Take note of this:

#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#

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Answer 2

William Abdalla

To evaluate ∫ ln(2x+1) dx, use integration by parts. Let u = ln(2x+1) and dv = dx. Then, differentiate u and integrate dv to find du and v respectively. After that, apply the integration by parts formula:

∫ u dv = uv - ∫ v du

Substitute the values of u, dv, du, and v into the formula and solve the integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.