What is #int (-x^3-2x-3 ) / (7x^4+ 5 x -1 )#?
1 Answer
Explanation:
What are the roots of the polynomial
This is a depressed quartic function and I recommend this 2 sources (the first is to show work, the second for a fast solution) to resolve it:
http://www.sosmath.com/algebra/factor/fac12/fac12.html
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
From the root finder we get:
#x_1=.19785#
#x_2=-.95254#
#x_3=.37734+i.78462#
#x_4=.37734-i.78462#
For not to deal with complex numbers we have
The original function can be rewritten in partial fractions in this way
For
#A/-.19785+B/.95254+0*C+D/.75468=3#
#A/-1.19785+B/.04746+(-C+D)/5.51269=0#
#A/.80215+B/1.95254+(C+D)/1.00333=-6/11#
#A/1.80295+B/2.95254+(2C+D)/3.24865=-15/121#
Or
#[[-5.05433,1.04982,0,1.31924],[-.83483,21.07038,-.39798,.39798],[1.24665,.51215,.99668,.99668],[.55489,.33869,.61564,.30782]][[A],[B],[C],[D]]=[[3],[0],[-6/11],[-15/121]]#
Solving this system of variables we get
#A=-.65697#
#B=-.01191#
#C=.51407#
#D=-.23349#
So the original expression becomes
#-.65697int dx/(x-.19785)-.01191int dx/(x+.95254)+int (.51407x-.23349)/(x^2-.75468x+.75801)dx# [#alpha# ]
Let's resolve the last part of the expression, the only one that poses a challenge
#(x-.37734)^2=x^2-.75468+.14239#
=>#x^2-.75468x+.75801=(x-.37734)^2+.61562#
#(x-.37734)=sqrt(.61562)tany=.78462tany#
#dx=.78462sec^2ydy#
How many units of#(x-.37734)# are there in the numerator?
#(.51407x-.23349)/(x-.37734)=.51407-.03951/(x-.37734)#
That's why now we are dealing with
#tany=(x-.37734)/.78462# =>#siny=(x-.37734)/.78462cosy#
=>#sin^2y+cos^2y=1# =>#((x^2-.75468x+.14239)/.61562+1)cos^2 y=1# =>#cosy=.78462/sqrt(x^2-.75468x+.75801)#
#-> =-.51407 ln |.78462/sqrt(x^2-.75468x+.75801)|-.05036tan^(-1)((x-.37734)/.78462)#
Therefore expression [