> 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

First execute long division and find the quotient and remainder.

> 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: .

The denominator is easily factored:

> factor(x^2-1);So, . Multiplying by the

> 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-2),x);To check the work let Maple do the integral directly.

int((5*x-1)/(x^2-1),x);Remember the constant: .

2011-02-17