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**

Prof. Neela, your answer is not correct. I calculated the derivative of you result and is not giving the function.

Look:

Your answer is (1/2) ln[(x-1/2)/(x+1)] .

The derivative is

(1/2) ln[(x-1/2)/(x+1)]'=(1/2)[(x+1)/(x-1/2)][1/2(x+1)^2]

(1/2) ln[(x-1/2)/(x+1)]'=2/4(2x-1)(x+1)

(1/2) ln[(x-1/2)/(x+1)]'=1/2(2x-1)(x+1), which is not the function f(x)=x/(2x+1)(x+1).

To calculate Integral xdx/[(2x+1)(x+1)]

=Integration xdx/(2x^2+3x+1) = xdx/[2(x^2+3/2x+1/2)]

=Integration xdx/2{(x+3/4)^2 - (3/4)^2+1/2}

=Integration xdx/{2(x+3/4)^2-1/16}

= Integration (1/4) dt/{t^2-(1/2)^2}, whre t = (x+3/4)^2 or xdx = dt/2. So,

=(1/4) dT/(T^2-A^2), wher T = t and A =1/4

=(1/4) (1/(2A)) ln(T-A)/T+A), as Int dx /(x^2-a^2) = (1/(2a))ln[(x-a)/x+a)] is a standard result of integration.

= (1/4)(2)ln{[(x+3/4)-1/4]/(x+3/4+1/4) + C

=(1/2) ln[(x-1/2)/(x+1)] + C , where C is a constant of integration