Given: #int 1/(2x+2x^2)dx =?#
Decompose the integrand into two fractions:
#1/(2x+2x^2) = A/(2x)+ B/(x+1)#
Multiply both sides by the denominator on the left:
#1 = A(x+1)+ B2x#
Let #x = 0#:
#1 = A(0+1)+ B2(0)#
#A = 1#
Let #x = -1#:
#1 = A(-1+1)+ B2(-1)#
#B = -1/2#
Check:
#1/(2x)-1/(2x+2)#
#1/(2x)(2x+2)/(2x+2)-1/(2x+2)(2x)/(2x)#
#(2x + 2 - 2x)/(4x+4x^2)#
#2/(4x+4x^2)#
#1/(2x+2x^2)#
This checks, therefore, we may write the following:
#int 1/(2x+2x^2)dx =int (1/(2x)-1/(2x+2)) dx#
This can be written as two integrals:
#int 1/(2x+2x^2)dx =1/2int 1/xdx -1/2int1/(x+1) dx#
These integrals are well known:
#int 1/(2x+2x^2)dx =1/2ln|x| -1/2ln|x+1|+ C#
#int 1/(2x+2x^2)dx =ln(sqrtx) -ln(sqrt(x+1))+ C#
#int 1/(2x+2x^2)dx =ln(sqrt(x/(x+1)))+ C#
