Just use the substitution you = x+1, then replace dx with du and you get you+1/you = 1 + 1/you which you can integrate to give you + ln you, thus = x-1 + ln(x-1)
How Do You Integrate (x/(x-1))?
I look it up in a table of integrals.
Integral[x dx/(ax+b)] = x/a - b/a2*ln[ax+b]
Your problem has a=1, b=-1
Integral[x dx/(x-1)] = x + ln[x-1]
Integral[x dx/(ax+b)] = x/a - b/a2*ln[ax+b]
Your problem has a=1, b=-1
Integral[x dx/(x-1)] = x + ln[x-1]
1st step you should know that f(x)= P(x)/Q(x) , Q(x)is not equal to zero.
2nd steps you already know that P(x)= x and Q(x)= x-1
3rd steps you should change x-1 equal ax + b to make easier when we solve I
4th steps you write integrate (x/(x-1)) = integrate (x/(ax + b))
5th steps you must time time the denominator both of side to become this ax^2 + bx = x^2 - x and compare it..you can get a = 1 and b = -1(try it carefully)*
then,the new formula is P(x) = aQ'(x) = b
solve it .
2nd steps you already know that P(x)= x and Q(x)= x-1
3rd steps you should change x-1 equal ax + b to make easier when we solve I
4th steps you write integrate (x/(x-1)) = integrate (x/(ax + b))
5th steps you must time time the denominator both of side to become this ax^2 + bx = x^2 - x and compare it..you can get a = 1 and b = -1(try it carefully)*
then,the new formula is P(x) = aQ'(x) = b
solve it .
