# Table of Integrals - Forms Involving ax + b

The integrals below involve ax + b, including forms where ax + b is raised to an exponent or in the denominator of a fraction.

1) int  1/(ax+b)  dx = 1/a ln(ax+b)

2) int  x/(ax+b)  dx = x/a - b/(a^2) ln(ax+b)

3) int  x^2/(ax+b)  dx = (ax+b)^2/(2a^3)-(2b(ax+b))/a^3+b^3/a^3 ln(ax+b)

4) int  x^3/(ax+b)  dx = (ax+b)^3/(3a^4)-(3b(ax+b)^2)/2a^4+(3b^2(ax+b))/a^4-b^3/a^4 ln(ax+b)

5) int  1/(x(ax+b))  dx = 1/b ln(x/(ax+b))

6) int  1/(x^2 (ax+b))  dx = -1/(bx)+a/b^2 ln((ax+b)/x)

7) int  1/(x^3(ax+b))  dx = (2ax-b)/(2b^2x^2)+a^2/b^3 ln(x/(ax+b))

8) int  1/(ax+b)^2  dx = -1/(a(ax+b))

9) int x/(ax+b)^2  dx = b/(a^2(ax+b))+1/a^2 ln(ax+b)

10) int x^2/(ax+b)^2  dx = (ax+b)/a^3-b^2/(a^3(ax+b))-(2b)/a^3 ln(ax+b)

11) int  x^3/(ax+b)^2  dx = (ax+b)^2/(2a^4)-(3b(ax+b))/a^4+b^3/(a^4(ax+b))+(3b^2)/a^4 ln(ax+b)

12) int  1/(x(ax+b)^2)  dx = 1/(b(ax+b))+1/b^2 ln(x/(ax+b))

13) int 1/(x^2(ax+b)^2)  dx = (-a)/(b^2(ax+b))-1/(b^2x)+(2a)/b^3 ln((ax+b)/x)

14) int 1/(x^3(ax+b)^2)  dx = -(ax+b)^2/(2b^4x^2)+(3a(ax+b))/(b^4x)-(a^3x)/(b^4(ax+b))-(3a^2)/b^4 ln((ax+b)/x)

15) int  1/(ax+b)^3  dx = (-1)/(2(ax+b)^2)

16) int  x/(ax+b)^3  dx = (-1)/(a^2(ax+b))+b/(2a^2(ax+b^2))

17) int  x^2/(ax+b)^3 = (2b)/(a^3(ax+b))-b^2/(2a^3(ax+b)^2)+1/a^3 ln(ax+b)

18) int  x^3/(ax+b)^3  dx = x/a^3-(3b^2)/(a^4(ax+b))+b^3/(2a^4(ax+b)^2)-(3b)/a^4 ln(ax+b)

19) int  1/(x(ax+b)^3)  dx = (a^2x^2)/(2b^3(ax+b)^2)-(2ax)/(b^3(ax+b))-1/b^3ln((ax+b)/x)

20) int  1/(x^2(ax+b)^3)  dx = (-a)/(2b^2(ax+b)^2)-(2a)/(b^3(ax+b))-1/(b^3x)+(3a)/b^4 ln((ax+b)/x)

21) int 1/(x^3(ax+b)^3)  dx = (a^4x^2)/(2b^5(ax+b)^2)-(4a^3x)/(b^5(ax+b)) - (ax+b)^2/(2b^5x^2)-(6a^2)/b^5 ln((ax+b)/x)

22) int  (ax+b)^n  dx = (ax+b)^(n+1)/((n+1)a), if n=-1, see (1)

23) int  x(ax+b)^n  dx = (ax+b)^(n+2)/((n+2)a^2)-(b(ax+b)^(n+1))/((n+1)a^2), n != -1, -2.  If n=-1, -2, See (2), (9)

24) int x^2(ax+b)^n dx = (ax+b)^(n+3)/((n+3)a^3)-(2b(ax+b)^(n+2))/((n+2)a^3)+(b^2(ax+b)^(n+1))/((n+1)a^3)

If n=-1, -2, -3, see (3), (10), (17)

25) int  x^m(ax+b)^n  dx = (x^(m+1)(ax+b^n))/(m+n+1)+(nb)/(m+n+1)int  x^m(ax+b)^(n-1)  dx

or: = (x^m(ax+b)^(n+1))/((m+n+1)a)-(mb)/((m+n+1)a)int  x^(m-1)(ax+b)^n  dx

or: =( -x^(m+1)(ax+b)^(n+1))/((n+1)b)+(m+n+2)/((n+1)b)int  x^m(ax+b)^(n+1)  dx`