Home > Calculus > Integration by Parts

How do you integrate #ln(2x+1)#?

Answer 1

Luke Alvarez

I would try by parts:

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

To integrate ( \ln(2x + 1) ), follow these steps:

Use integration by parts: [ \int \ln(u) , du = u \ln(u) - \int \frac{du}{u} ]
Let ( u = 2x + 1 ) and ( du = 2 , dx ).
Substitute ( u = 2x + 1 ) and ( du = 2 , dx ) into the integral: [ \int \ln(2x + 1) , dx = \int \ln(u) , du ]
Integrate ( \int \ln(u) , du ) using the integration by parts formula.
Substitute back ( u = 2x + 1 ) to get the final result.

By signing up, you agree to our Terms of Service and Privacy Policy

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.