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)

I look it up in a table of integrals.

Integral[x dx/(ax+b)] = x/a - b/a

Your problem has a=1, b=-1

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

Integral[x dx/(ax+b)] = x/a - b/a

^{2}*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 .