What is the integration of #(x^6+1)/(x+1)# ?

2 Answers
George C.
Mar 23, 2018

#int (x^6+1)/(x+1) dx = 1/6x^6-1/5x^5+1/4x^4-1/3x^3+1/2x^2-x + 2ln abs(x+1) + C#

Explanation:

Note that:

#x^6 + 1 = (x+1)(x^5-x^4+x^3-x^2+x-1)+2#

So:

#int (x^6+1)/(x+1) dx = int x^5-x^4+x^3-x^2+x-1 + 2/(x+1) dx#

#color(white)(int (x^6+1)/(x+1) dx) = 1/6x^6-1/5x^5+1/4x^4-1/3x^3+1/2x^2-x + 2ln abs(x+1) + C#

Sahar Mulla ❤
Mar 23, 2018

#int (x^6+1)/(x+1)dx#

#=> int(((x+1)(x^5-x^4+x^3_x^2+x) + (-x+1))/(x+1))dx#

#=>int( (cancel((x+1))(x^5-x^4+x^3-x^2+x))/cancel(x+1) + (-x+1)/((x+1)))dx#

#=>int(x^5-x^4+x^3-x^2+x)dx + int(1-x)/(x+1)dx#

#=>int(x^5-x^4+x^3-x^2+x)dx + int1/(x+1)dx-intx/(x+1)dx#

Usig the rule, #color(red)(intx^n dx= x^(n+1)/(n+1)# and #color(red)(int 1/x dx= log|x|#

#=>(x^6/6-x^5/5+x^4/4-x^3/3+x^2/2) + log|x+1|+C-intx/(x+1)dx#

#color(white)(dddd#

#color(white)(dddd#

Solving, #intx/(x+1)dx#

Let, #x+1 = t => dx=dt#

#:. intx/(x+1)dx = int(t-1)/tdt#

#=> int1-1/tdt #

#=> t -log|t|#

#=> x+1 - log|x+1|# [Replacing, #t=x+1# back.]

#color(white)(dddd#

#color(white)(dddd#

Final answer:-

#=>(x^6/6-x^5/5+x^4/4-x^3/3+x^2/2) + log|x+1|-(x+1) + log|x+1|#

#=>(x^6/6-x^5/5+x^4/4-x^3/3+x^2/2) + log|x+1|^2-(x+1)+C#

#=>x^6/6-x^5/5+x^4/4-x^3/3+x^2/2-x + log|x+1|^2+1+ C#

#=>x^6/6-x^5/5+x^4/4-x^3/3+x^2/2-x + log|x+1|^2+C# (note that, #1+C# is taken as #C# as it is a constant,)

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