> 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: .

Make sure to show your steps and include plenty of text to keep your work clear. Also check your final answer by having Maple do the integral directly.

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: .

Make sure to show your steps and include plenty of text to keep your work clear. Also check your final answer by having Mapel do the integral directly.

2006-02-13