Collective Table of Formulas

Indefinite Integrals with ax+b

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$ \int \frac {d x}{ ax + b} = \frac {1}{a} \ln (ax +b)+C $
$ \int \frac {x d x}{ ax + b} = \frac {x}{a} - \frac{b}{a^2} \ln (ax +b)+C $
$ \int \frac {x^2 d x}{ ax + b} = \frac {(ax+b)^2}{2a^3} - \frac {2b(ax+b) }{a^3} + \frac{b^2}{a^3} \ln (ax +b)+C $
$ \int \frac {x^3 d x}{ ax + b} = \frac {(ax+b)^3}{3a^4} - \frac {3b(ax+b)^2 }{2a^4} + \frac{3b^2(ax+b)}{a^4} - \frac{b^3}{a^3}\ln (ax +b)+C $
$ \int \frac {d x}{ x(ax + b)} = \frac {1}{b} \ln \left ( \frac {x}{ax +b} \right)+C $
$ \int \frac {d x}{ x^2(ax + b)} = - \frac {1}{b x} + \frac {a}{b^2} \ln \left ( \frac {ax +b}{x} \right)+C $
$ \int \frac {d x}{ x^3(ax + b)} = \frac {2 a x - b}{2 b^2 x^2} + \frac {a^2}{b^3} \ln \left ( \frac {x}{ax+b} \right)+C $
$ \int \frac {d x}{(ax + b)^2} = \frac {-1}{a(ax+b)} +C $
$ \int \frac {x d x}{(ax + b)^2} = \frac {b}{a^2(ax+b)} + \frac {1}{a^2} \ln (ax+b) +C $
$ \int \frac {x^2 d x}{(ax + b)^2} = \frac {ax+b}{a^3} - \frac{b^2}{a^3(ax+b)} - \frac {2b}{a^3} \ln (ax+b) +C $
$ \int \frac {x^3 d x}{(ax + b)^2} = \frac {(ax+b)^2}{2a^4} - \frac {3b(ax+b)}{a^4} +\frac{b^3}{a^4(ax+b)} + \frac {3b^2}{a^4} \ln (ax+b) +C $
$ \int \frac {d x}{x(ax + b)^2} = \frac {1}{b(ax+b)} + \frac {1}{b^2} \ln \left ( \frac{x}{ax+b} \right ) +C $
$ \int \frac {d x}{x^2(ax + b)^2} = \frac {-a}{b^2(ax+b)} - \frac {1}{b^2x} + \frac {2a}{b^3} \ln \left ( \frac {ax+b}{x} \right ) +C $
$ \int \frac {d x}{x^3(ax + b)^2} = - \frac {(ax+b)^2}{2b^4x^2} + \frac {3 a(ax+b)}{b^4x} - \frac {a^3 x}{b^4(ax+b)} - \frac{3a^2}{b^4} \ln \left ( \frac {ax+b}{x} \right ) +C $
$ \int \frac {d x}{(ax + b)^3} = \frac {-1}{2(ax+b)^2} +C $
$ \int \frac {x d x}{(ax + b)^3} = \frac {-1}{a^2(ax+b)} + \frac {b}{2a^2(ax+b)^2} +C $
$ \int \frac {x^2 d x}{(ax + b)^3} = \frac {2b}{a^3(ax+b)} - \frac {b^2}{2a^3(ax+b)^2} + \frac {1}{a^3} \ln (ax+b) +C $
$ \int \frac {x^3 d x}{(ax + b)^3} = \frac {x}{a^3} - \frac {3b^2}{a^4(ax+b)} + \frac {b^3}{2a^4(ax+b)^2} - \frac {3b}{a^4} \ln (ax+b) +C $
$ \int \frac {d x}{x(ax + b)^3} = \frac {a^2x^2}{2b^3(ax+b)^2} - \frac {2ax}{b^3(ax+b)} - \frac {1}{b^3} \ln \left( \frac{ax+b}{x} \right) +C $
$ \int \frac {d x}{x^2(ax + b)^3} = \frac {-a}{2b^2(ax+b)^2} - \frac {2a}{b^3(ax+b)} - \frac {1}{b^3x} + \frac {3a}{b^4} \ln \left( \frac{ax+b}{x} \right) +C $
$ \int \frac {d x}{x^3(ax + b)^3} = \frac {a^4x^2}{2b^5(ax+b)^2} - \frac {4a^3x}{b^5(ax+b)} - \frac {(ax+b)^2}{2b^5x2} - \frac {6a^2}{b^5} \ln \left( \frac{ax+b}{x} \right) +C $
$ \int (a x +b)^n d x = \frac {(ax+b)^{n+1} }{(n+1)a}+C. \qquad n =-1 $
$ \int x (a x +b)^n d x = \frac {(ax+b)^{n+2} }{(n+2)a^2} - \frac {b(ax+b)^{n+1}}{(n+1)a^2}+C, \qquad n \neq -1,-2 $
$ \int x^2 (a x + b)^n d x = \frac {(ax+b)^{n+3} }{(n+3)a^3} - \frac {2b(ax+b)^{n+2}}{(n+2)a^3} + \frac {b^2(ax+b)^{n+1}}{(n+1)a^3}+C, \qquad n = -1,-2, -3 $
$ \int x^m (a x + b)^n d x = \begin{cases} \frac {x^{m+1}(ax+b)^n}{m + n + 1} + \frac {n b}{m + n+ 1} \int x^m (ax+b)^{n-1} d x \\ \frac {x^m(ax+b)^{n+1}}{(m + n + 1)a} - \frac {m b}{(m + n+ 1)a} \int x^{m-1} (ax+b)^{n} d x \\ \frac {- x^{m+1}(ax+b)^{n+1}}{(n + 1)b} + \frac {m+ n+ 2 }{(n+ 1)b} \int x^m (ax+b)^{n+1} d x \end{cases} $
$ \int \frac {d x}{\sqrt{a x +b}} = \frac {2\sqrt{ax+b}}{a}+C $
$ \int \frac {x d x}{\sqrt{a x + b}} = \frac {2(ax-2b)}{3a^2}\sqrt{ax+b}+C $
$ \int \frac {x^2 d x}{\sqrt{a x + b}} = \frac {2(3a^2x^2-4abx + 8b^2)}{15a^3}\sqrt{ax+b}+C $
$ \int \frac {d x}{x \sqrt {ax+b}} = \begin{cases} \frac {1}{b} \ln \left ( \frac {\sqrt {ax+b} - \sqrt {b}}{\sqrt {ax+b} + \sqrt {b}} \right ) +C\\ \frac {2}{\sqrt {-b}} \arctan \sqrt { \frac {ax+b}{- b}} +C \\ \end{cases} $
$ \int \frac { d x}{x ^2 \sqrt{a x + b}} = - \frac {\sqrt{ax+b}}{b x} - \frac {a}{2 b} \int \frac {d x}{x \sqrt {ax + b}}+C $
$ \int \sqrt{a x + b} \ d x = \frac {2 \sqrt{(ax+b)3}}{3 a}+C $
$ \int x \sqrt{a x + b} \ d x = \frac {2(3ax-2b)}{15a^2}\sqrt{(ax+b)^3}+C $
$ \int x^2 \sqrt{a x + b} \ d x = \frac {2(15a^2x^2-12abx + 8b^2)}{105a^3}\sqrt{(ax+b)^3}+C $
$ \int \frac {\sqrt {ax+b}}{x} \ d x = 2 \sqrt {ax+b} + b \ \int \frac {d x}{x \sqrt {ax + b}} +C $
$ \int \frac {\sqrt {ax+b}}{x^2} \ d x = - \frac {\sqrt {ax+b}}{x} + \frac {a}{2} \int \frac {d x}{x \sqrt {ax + b}} +C $
$ \int \frac {x^m}{\sqrt{ ax+b}} d x = \frac {2x^m \sqrt {ax+b}}{(2m+1)a} - \frac {2mb}{(2m+1)a} \int \frac {x^{m-1}}{\sqrt {ax+b}} d x+C $
$ \int \frac {d x}{x^m \sqrt{ax+b}} =- \frac {\sqrt{ax+b}}{(m-1)bx^{m-1}} - \frac {(2m-3)a}{(2m-2)b} \int \frac {d x}{x^{m-1} \sqrt{ax+b}}+C $
$ \int x^m \sqrt {ax+b} \ d x = \frac{2x^m}{(2m+3)a}(a+b)^{\frac{3}{2}} -\frac {2mb}{(2m+3)a} \int x^{m-1} \sqrt{ax+b} \ d x+C $
$ \int \frac {\sqrt {ax+b}}{x^m} d x = - \frac {\sqrt {ax+b}}{(m-1)x^{m-1}} + \frac {a}{2(m-1)} \int \frac {d x}{x^{m-1} \sqrt {ax+b}} +C $
$ \int \frac {\sqrt {ax+b}}{x^m} d x = \frac {-(ax+b)^{3/2}}{(m-1)bx^{m-1}} - \frac {(2m-5)a}{(2m-2)b} \int \frac {\sqrt {ax+b}}{x^{m-1}} d x +C $
$ \int (ax+b)^{m/2} d x = \frac {2(ax+b)^{(m+2)/2}}{a(m+2)}+C $
$ \int x(ax+b)^{m/2} d x = \frac {2(ax+b)^{(m+4)/2}}{a^2(m+4)} - \frac {2b(ax+b)^{(m+2)/2}}{a^2(m+2)}+C $
$ \int x^2(ax+b)^{m/2} d x = \frac {2(ax+b)^{(m+6)/2}}{a^3(m+6)} - \frac {4b(ax+b)^{(m+4)/2}}{a^3(m+4)}+ \frac {2b^2(ax+b)^{(m+2)/2}}{a^3(m+2)}+C $
$ \int \frac {(ax+b)^{m/2}}{x} d x =\frac {2(ax+b)^{m/2}}{m} + b \ \int \frac {(ax+b)^{(m-2)/2}}{x} d x $
$ \int \frac {(ax+b)^{m/2}}{x^2} d x = - \frac {(ax+b)^{(m+2)/2}}{bx} + \frac {ma}{2b} \ \int \frac {(ax+b)^{m/2}}{x} d x $
$ \int \frac {d x}{x(ax+b)^{m/2}} d x = \frac {2}{(m-2)b(ax+b)^{(m-2)/2}} + \frac {1}{b} \ \int \frac {d x}{x(ax+b)^{(m-2)/2}} $

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Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood