# Ostobee Zorn  Chapter 9
# 
# 9.1   #7   The hint is not needed.
# 
> with(student);
Warning, new definition for D

[D, Diff, Doubleint, Int, Limit, Lineint, Product, Sum, Tripleint,

    changevar, combine, completesquare, distance, equate, extrema,

    integrand, intercept, intparts, isolate, leftbox, leftsum,

    makeproc, maximize, middlebox, middlesum, midpoint, minimize,

    powsubs, rightbox, rightsum, showtangent, simpson, slope,

    trapezoid, value]

> F := Int(x*cos(x)^2,x);

                               /
                              |          2
                        F :=  |  x cos(x)  dx
                              |
                             /

> F = intparts(F,x);

  /
 |          2
 |  x cos(x)  dx =
 |
/

                                      /
                                     |
    x (1/2 sin(x) cos(x) + 1/2 x) -  |  1/2 sin(x) cos(x) + 1/2 x dx
                                     |
                                    /

# Now we can  integrate by inspection.
>  F = x*(1/2*sin(x)+1/2*x)-1/2*(1/2*sin(x)^2 + 1/2*x^2) + C;

   /
  |          2                                         2        2
  |  x cos(x)  dx = x (1/2 sin(x) + 1/2 x) - 1/4 sin(x)  - 1/4 x  + C
  |
 /

# checking,
> int(x*cos(x)^2,x);

                                                   2        2
         x (1/2 sin(x) cos(x) + 1/2 x) + 1/4 cos(x)  - 1/4 x

# explain the difference.
# 
# #12    
> F:=Int(x*ln(x),x ); 

                                /
                               |
                         F :=  |  x ln(x) dx
                               |
                              /

> intparts(F,ln(x));

                                      /
                                2    |
                     1/2 ln(x) x  -  |  1/2 x dx
                                     |
                                    /

# So, now we can write an antiderivative for xln(x)
> F := x -> 1/2*ln(x)*x^2 - 1/4*x^2;

                                        2        2
                   F := x -> 1/2 ln(x) x  - 1/4 x

# So the definite integral evaluated by the fundamental thm is
> F(exp(1))-F(1);

                                    2
                          1/4 exp(1)  + 1/4

# Checking this directly.
> Int(x*ln(x),x=1..exp(1))=int(x*ln(x),x=1..exp(1));

                  exp(1)
                 /
                |
                |        x ln(x) dx = 1/4 exp(2) + 1/4
                |
               /
                 1

> 
# 
#  9.2  
# 
# # 5  
# 
>  f :=  (x^2-1)/(x*(x^2+4));

                                       2
                                 -1 + x
                           f := ----------
                                    2
                                x (x  + 4)

> convert(f,parfrac,x);

                                          x
                        - 1/4 1/x + 5/4 ------
                                         2
                                        x  + 4

#  We can integrate each part by inspection.
>  F := -1/4*ln(x) + 5/4* 1/2*ln(x^2+4) +C;

                                           2
                F := - 1/4 ln(x) + 5/8 ln(x  + 4) + C

# Checking this
> Int((x^2-1)/(x*(x^2+4)),x)=int((x^2-1)/(x*(x^2+4)),x);

             /        2
            |   -1 + x                               2
            |  ---------- dx = - 1/4 ln(x) + 5/8 ln(x  + 4)
            |      2
           /   x (x  + 4)

# #17  This is the one we did in class on Monday.
> restart: with(student):f := x^2/((x^2+1) *(x+1)^2);

                                     2
                                    x
                        f := -----------------
                               2             2
                             (x  + 1) (x + 1)

# The form for the partial fraction decomposition of f is
>  g := (a*x+b)/(x^2+1) + c/(x+1) + d/(x+1)^2;

                        a x + b     c        d
                   g := ------- + ----- + --------
                         2        x + 1          2
                        x  + 1            (x + 1)

> collect(simplify(g),x);

            3                      2
   (a + c) x  + (d + b + 2 a + c) x  + (2 b + a + c) x + b + d + c
   ---------------------------------------------------------------
                            2             2
                          (x  + 1) (x + 1)

> solve({a+c=0,d+b+2*a+c=1,2*b+a+c=0,b+d+c=0},{a,b,c,d});

                 {d = 1/2, a = 1/2, c = -1/2, b = 0}

> assign(");
> g;

                      x            1            1
                1/2 ------ - 1/2 ----- + 1/2 --------
                     2           x + 1              2
                    x  + 1                   (x + 1)

> f := convert(f,parfrac,x);

                          1             1           x
              f := 1/2 -------- - 1/2 ----- + 1/2 ------
                              2       x + 1        2
                       (x + 1)                    x  + 1

#  We can integrate the parts by inspection.
> F := -1/2*1/(x+1)-1/2*ln(x+1) + 1/4*ln(x^2+1)+C;

                     1                             2
        F := - 1/2 ----- - 1/2 ln(x + 1) + 1/4 ln(x  + 1) + C
                   x + 1

# Checking with Maple,
> int(f,x);

                     1                             2
             - 1/2 ----- - 1/2 ln(x + 1) + 1/4 ln(x  + 1)
                   x + 1

> f := x->x*exp(-x);

                         f := x -> x exp(-x)

> int(f(x),x=0..infinity);

                                  1

> plot(f,0..7);

> fsolve(int(f(x),x=0..z)=1/2,z);

                             1.678346990

>