Expert: Alon Mandes - 12/16/2008

Question
I need to find the following indefinite integrals and am having trouble.  Can you please help?  I'm a mom and my daughter thinks I can help her with this.

sec x tan x
____________dx
sex + 1


x to the third power - 6x - 20
_______________________________dx (hint: use long division
         x + 5                          to rewrite the
                                        expression in the
                                        form of q(x)+r(x)/
                                        d(x)
                                        r(x) = remainder
                                        d(x) = divisor

Answer
I'll start from the second :

(x^3-6x-20)/(x+5)= x^3/(x+5)- 6x/(x+5) - 20/(x+5).

1.
x^3/(x+5)=(x^3+5x^2-5x^2)/(x+5)=1-5x^2/(x+5)=
1+(-5x^2+x-x)/(x+5)=1+ (-5x^2-x)/(x+5)+ x/(x+5)=
1-x+ x/(x+5)=1-x+ x/(x+5)=1-x+(x+5-5)/(x+5)=1-x+1-5/(x+5)=

->∫x^3/(5+x) dx ∫x^3 dx= ∫(2-x)-5∫1/(x+5) dx =2x-(x^2)/2-5Ln(x+5).

2.
x/(x+5)=(x+1-1)/(x+5)=1-1/(x+5)

->∫6x/(x+5) dx = 6∫1-1/(x-5) dx=6x-6Ln(x+5).

3.
∫20/(x+5) dx = 20Ln(x+5).   SO,

∫(x^3-6x-20)/(x+5)=2x-(x^2)/2-5Ln(x+5)-6x+6Ln(x+5)-20Ln(x+5).


The first one :
(secx*tanx)/(1+secx)=tanx+sinx. I'll prove it to you:
keep in mind that tanx=sinx/cosx & secx=1/cosx. Then

sinx
--------       =
cosx^2
--------    
1/cosx +1


sinx
--------       
cosx^2
--------      =    
cosx+1
--------
cosx


sinx
--------       
cosx^2            sinx     cosx
--------      =  --------* -----    =
cosx+1            cosx^2   1+cosx
--------
cosx



sinx
----  *  (1+cosx)   =  tanx+sinx
cosx

∫sinx=cosx  & ∫tanx=1/cosx^2.

Alon.