One: the formula of indefinite integral (all in basic mathematics books, take notes carefully and record it yourself. )

∫ a dx = ax+C, a and c are constants.

∫ x a dx = [x (a+1)]/(a+1)+c, where a is a constant and a≦- 1.

∫ 1/x dx = ln|x| + C

∫ a x dx = (1/lna) a x+c, where a >；; 0 and a ≠ 1

* e^x dx = e^x+c

∫ cosx dx = sinx + C

∫ sinx dx = - cosx + C

∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

∫ tanx dx = - ln|cosx| + C = ln|secx| + C

∫secx dx = ln | cot(x/2)|+C =( 1/2)ln |( 1+sinx)/( 1-sinx)|+C =-ln | secx-tanx |+C = ln | secx+tanx |+C

∫cscx dx = ln | tan(x/2)|+C =( 1/2)ln |( 1-cosx)/( 1+cosx)|+C =-ln | cscx+cotx |+C = ln | cscx-cotx |+C

∫ sec^2(x) dx = tanx + C

∫ csc^2(x) dx = - cotx + C

∫ secxtanx dx = secx + C

∫ cscxcotx dx = - cscx + C

∫ dx/(a 2+x 2) = (1/a) arctangent (x/a)+C

∫ dx/√ (a 2-x 2) = arcsine (x/a)+C

∫dx/√(x^2+a^2)= ln | x+√(x^2+a^2)|+c

∫dx/√(x^2-a^2)= ln | x+√(x^2-a^2)|+c

∫√(x^2-a^2)dx =(x/2)√(x^2-a^2)-(a^2/2)ln|x+√(x^2-a^2)|+c

∫√(x^2+a^2)dx =(x/2)√(x^2+a^2)+(a^2/2)ln|x+√(x^2+a^2)|+c

∫√(a^2-x^2)dx =(x/2)√(a^2-x^2)+(a^2/2)arcsin(x/a)+c

Two: What are the basic formulas of indefinite integral?

Three: What is indefinite integral?

If f(x) is the derivative function of f(x), then F(x)+C(C is an arbitrary constant) is the indefinite integral of F(x), and the indefinite integral of F(x) is expressed as ∫ f (x) dx = f.c. 。