> diff(x^2+2*x+9,x);Note that the numerator times a constant is the derivative of the denominator.
> simplify((6*x+6)/(2*x+2));Therefore, which is an easy natural log integral.
> int(3/u,u); > subs(u=x^2+2*x+9,int(3/u,u));To check the work, let Maple do the intgral directly.
> int((6*x+6)/(x^2+2*x+9),x);Remember with indefinite integrals the solution adds a constant. So, the inetgral solution is . Often a function is not in that straight forward form. With long division, you can try and get the quotient function into the form of a polynomial plus a fraction where the numerator is a derivative of the denominator: . For example if
> q:=quo((x^3+x^2+x-1),(x^2+2*x+2),x); > r:=rem((x^3+x^2+x-1),(x^2+2*x+2),x);The new form of the function is:
> f:=q+r/(x^2+2*x+2);Note that the fractional part has the numerator a derivative times a constant of the denominator.
> diff(x^2+2*x+2,x); > simplify(r/diff(x^2+2*x+2,x)); > int(q,x)+subs(u=x^2+2*x+2,int(1/(2*u),u));To check the work, let Maple do the integral directly.
> int((x^3+x^2+x-1)/(x^2+2*x+2),x);Remember to add a constant to the indefinite integral answer: .
> factor(x^2-1);So, . Multiplying by the common denominator and expanding gives:
> expand(5*x-1=A*(x-1)+B*(x+1));With this equation we can solve for and by equating the coefficients of the x term and then equating the constants. This will give us two equations which can be solved simultaneously.
> solve({5*x=A*x+x*B,-1=-A+B},{A,B});These values tell us that: . The right-hand side shows fractions that are easily integrated with the natural log.
> int(3/(x+1)+2/(x-1),x);To check the work let Maple do the integral directly.
int((5*x-1)/(x^2-1),x);Remember the constant: .