What is #int (-x^3+9x-1 ) / (-2x^2- 3 x +5 )#?

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Answer 1 · 2016-12-08T11:54:11

The answer is #=x^2/4-3/4x-9/8ln(∣2x+5∣)-ln(∣x-1∣)+C#

Let's do a long division

#color(white)(aaaa)##-x^3+##color(white)(aaaaaaaaa)##9x-1##color(white)(aa)##∣##-2x^2-3x+5#

#color(white)(aaaa)##-x^3+##color(white)(a)##-3/2x^2+5/2x##color(white)(aaaaa)##∣##x/2-3/4#

#color(white)(aaaaaaa)##0+##color(white)(a)##+3/2x^2+13/2x-1#

#color(white)(aaaaaaaaaaa)####color(white)(aaa)##3/2x^2+9/4x-15/4#

#color(white)(aaaaaaaaaaa)####color(white)(aaaaa)##0+17/4x+11/4#

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

So,

#(-x^3+9x-1)/(-2^2-3x+5)=(x/2-3/4)+(17/4x+11/4)/(-2x^2-3x+5)#

#=(x/2-3/4)-(17/4x+11/4)/((2x+5)(x-1))#

Let's do a partial fraction decomposition

#(17/4x+11/4)/((2x+5)(x-1))=A/(2x+5)+B/(x-1)#

#=(A(x-1)+B(2x+5))/((2x+5)(x-1))#

#17/4x+11/4=A(x-1)+B(2x+5)#

Let #x=1#, #=>#, #28/4=7B#, #=>#, #B=1#

Let #x=0#,#=>#,#11/4=-A+5B#

#A=5-11/4=9/4#

So,
#int((-x^3+9x-1)dx)/(-2^2-3x+5)#

#=int(x/2-3/4)dx-9/4intdx/(2x+5)-intdx/(x-1)#

#=x^2/4-3/4x-9/8ln(∣2x+5∣)-ln(∣x-1∣)+C#