SOLUTION: Evaluate the integral using integration by parts with explanations. ln (2x+1) dx

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Question 407767: Evaluate the integral using integration by parts with explanations.
ln (2x+1) dx
Found 2 solutions by Tatiana_Stebko, richard1234:
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
int%28ln%282x%2B1%29dx%29
Use formula int%28udv%29=u%2Av-int%28vdu%29
Let u=ln%282x%2B1%29 dv=dx
then du=%282%2F%282x%2B1%29%29dx v=x
int%28ln%282x%2B1%29dx%29=xln%282x%2B1%29-int%28x%2A%282%2F%282x%2B1%29%29dx%29=xln%282x%2B1%29-int%28%28%282x%29%2F%282x%2B1%29%29dx%29=xln%282x%2B1%29-int%28%28%282x%2B1%29-1%29%2F%282x%2B1%29%29dx%29=xln%282x%2B1%29-int%28%281-%281%2F%282x%2B1%29%29%29dx%29=xln%282x%2B1%29-%28x-2%2F%282x%2B1%29%29%2BC=xln%282x%2B1%29-x%2B2%2F%282x%2B1%29%2BC

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
This is an alternative approach which can still use integration by parts, but is slightly easier.

Let u+=+2x%2B1, du+=+2dx, dx+=+du%2F2. If we replace 2x+1, we obtain

%281%2F2%29int%28ln%28u%29%2C+du%29. We can integrate int%28ln%28u%29%2C+du%29 via parts, or by a formula which you can (and should) memorize.

Letting y+=+ln+%28u%29, dz+=+du, we obtain dy+=+du%2Fu and z+=+u. Applying the parts formula,

int%28ln%28u%29%2C+du%29+=+u%2Aln+%28u%29+-+int%28u%2Fu%2C+du%29+=+u%2Aln+%28u%29+-+u+%2B+C. Note that we have to multiply by 1/2, since we want %281%2F2%29int%28ln%28u%29%2C+du%29. Back-substitute u+=+2x%2B1 to obtain