Calculate Integral xdx/(2x+1)(x+1).

We have to write the function as a sum of simple
ratio:

x/(x+1)(2x+1) = A/(x+1) +
B/(2x+1)

Bringing the 2 ratio to the same denominator,
we'll have:

x = A(2x+1) +
B(x+1)

x = 2Ax + A + Bx +
B

x = x(2A+B) + A + B

Being an
identity, the expression above has to have the corresponding coefficients
equal.

So, the coefficient of x, from the left side, is 0,
the left side being written: x+0.

According to
this:

2A+B=1

A + B=0, so
A=-B

-2B+B=1

-B=1

B=-1,
so A=1

x/(x+1)(2x+1) = 1/(x+1) -
1/(2x+1)

Integral [xdx/(x+1)(2x+1)]=Integral
dx/(x+1)-Integral[dx/(2x+1)]

Integral dx/(x+1) =
ln(x+1)

-Integral [dx/(2x+1)] =
-ln(2x+1)

Integral [xdx/(x+1)(2x+1)] = ln(x+1)-ln(2x+1) +
C

Integral [xdx/(x+1)(2x+1)] =
ln[(x+1)/(2x+1)] + C