use integration by parts to find the integral of ln(x+1) dx
any help would be appreciated
Whoah! Careful now.
You could split
$\displaystyle
\frac{x}{x+1}
$
into
$\displaystyle
1 - \frac{1}{x+1}
$
and have
$\displaystyle
I = x\ \ln(x+1) - \int (\ 1 - \frac{1}{1+x}\ )\ dx
$
$\displaystyle
\ \ =x\ \ln(1+x) - x + \ln(x+1)+C
$
which is fine.
But here's a short cut - notice that one integral of 1 is x + 1, and choose u = x+1 for du = 1.
I.e.
$\displaystyle
I = (x+1)\ \ln(x+1) - \int (x+1)\ \frac{1}{x+1}\ dx
$
$\displaystyle
= (x+1)\ \ln(x+1) - \int (1)\ dx
$
$\displaystyle
\ \ = (x+1)\ \ln(x+1) - x + C
$
Picture shortly...
Don't integrate - balloontegrate!
Balloon Calculus: worked examples from past papers