To find:
#int(-2x^3-x)/(-4x^2+2x+7)dx#
Let #I=int(-2x^3-x)/(-4x^2+2x+7)dx#
Dividing the fraction in the integrand
We get
#(-4x^2+2x+7)xx0.5x=-2x^3+x^2+3.5#
#-2x^3-x=-2x^3+x^2+3.5-x^2-3.5-x#
Thus,
#I=int(-2x^3+x^2+3.5-x^2-3.5-x)/(-4x^2+2x+7)dx#
#int(-2x^3+x^2+3.5)/(-4x^2+2x+7)dx+int(-x^2-3.5-x)/(-4x^2+2x+7)dx#
Ler #I_1=int(-2x^3+x^2+3.5)/(-4x^2+2x+7)dx and I_2=int(-x^2-3.5-x)/(-4x^2+2x+7)dx#
Then,
#I=I_1+I_2#
#I_1=int(-2x^3+x^2+3.5)/(-4x^2+2x+7)dx=int0.5dx=0.5x#
#I_1=0.5x#
#I_2=int(-x^2-3.5-x)/(-4x^2+2x+7)dx#
Taking negative sign, and rearranging the numerator
#I_2=int(x^2+x+3.5)/(4x^2-2x-7)dx#
Now, the numerator can be expressed as a combination
#(x^2+x+3.5)=p(4x^2-2x-7)+qd/dx(4x^2-2x-7)+r#
#d/dx(4x^2-2x-7)=8x-2#
Thus,
#x^2+x+3.5=p(4x^2-2x-7)+q(8x-2)+r#
Simplifying
#x^2+x+3.5=4px^2-2px-7p+8qx-2q+r#
Collecting the terms having like powers of x
#x^2+x+3.5=4px^2+(-2p+8q)x+(-7p-2q+r)#
Equating the coefficients of like powers of x
#1=4p#
#p=1/4=0.25#
#1=-2p+8q#
#1=-2xx1/4+8q#
#1=-1/2+8q#
#3/2=8q#
#q=3/16=0.1875#
#3.5=-7p-2q+r#
#3.5=-7xx0.25-2xx0.1875+r#
#3.5=-1.75-0.375+r#
#3.5=-2.125+r#
#r=3.5+2.125#
#r=5.625#
Thus, we have,
#p=0.25, q=0.1875, and r=5.625#
#p(4x^2-2x-7)+q(8x-2)+r=0.25(4x^2-2x-7)+0.1875(8x-2)+5.625#
and
#int(x^2+x+3.5)/(4x^2-2x-7)dx=int(0.25(4x^2-2x-7)+0.1875(8x-2)+5.625/(4x^2-2x-7))dx#
Using the sum rule and simplifying
#I_2=0.25int(4x^2-2x-7)/(4x^2-2x-7)dx+0.1875int((8x-2)dx)/(4x^2-2x-7)+5.625int1/(4x^2-2x-7)dx#
#0.25int(4x^2-2x-7)/(4x^2-2x-7)dx=0.25int1dx=0.25x#
#0.1875int(2x-2)/(4x^2-2x-7)dx=#
Let #t=4x^2-2x-7, dt=(8x-2)dx#
Then
#0.1875int(2x-2)/(4x^2-2x-7)dx=0.1875int(dt)/t=0.1875lnt#
Substituting for t
#0.1875int(2x-2)/(4x^2-2x-7)dx=0.1875ln(4x^2-2x-7)#
#5.625int1/(4x^2-2x-7)dx=#
Completing the squares in the denominator
#4x^2-2x-7=4(x^2-2/4x-7/4)#
#=4(x^2-2xx1/4x+(1/4)^2-7/4-(1/4)^2)#
#=4((x-1/4)^2-(7/4+1/16))#
#=4((x-1/4)^2-1.8125)#
#1.8125=1.346^2#
Thus,
#5.625int1/(4x^2-2x-7)dx=5.625int1/(4((x-1/4)^2-1.346^2))dx#
#Let t=x-1/4, dt=dx#
Now,
#5.625int1/(4x^2-2x-7)dx=5.625/4int(dt)/(t^2-1.346^2)dx#
#5.625/4=1.40625#
#int(dt)/(t^2-1.346^2)=int(dt)/((t+1.346)(t-1.346))#
#int(dx)/((x+a)(x+b))=1/(b-a)ln((a+x)/(b+x)), aneb#
We have,
#x=t, a=1.346, b=-1.346, b-a=-1.346-1.346=-2.692#
Thus,
#int(dt)/((t+1.346)(t-1.346))=1/(-2.692)ln((1.346+t)/(-1.346+t))#
Substituting for t
#1.346+t=1.346+(x-1/4)=1.346+x-0.25=x+1.096#
#-1.346+t=-1.346+(x-1/4)=-1.346+x-0.25=x-1.596#
#int(dt)/((t+1.346)(t-1.346))=1/(-2.692)ln((x+1.096)/(x-1.596))#
Now,
#5.625int1/(4x^2-2x-7)dx#
#=1.40625xx1/(-2.692)ln((x+1.096)/(x-1.596))#
#=-1.637ln((x+1.096)/(x-1.596))#
Thus,
#I_1=0.5x#
#I_2=int(-x^2-3.5-x)/(-4x^2+2x+7)dx#
#=0.25x#
#+0.1875ln(4x^2-2x-7)#
#+(-1.637ln((x+1.096)/(x-1.596)))#
#I_2=0.25x+0.1875ln(4x^2-2x-7)-1.637ln((x+1.096)/(x-1.596))#
I=I_1+I_2
#int(-2x^3-x)/(-4x^2+2x+7)dx#
#=0.5x+0.25x+0.1875ln(4x^2-2x-7)-1.637ln((x+1.096)/(x-1.596))#
#int(-2x^3-x)/(-4x^2+2x+7)dx#
#=0.75x+0.1875ln(4x^2-2x-7)-1.637ln((x+1.096)/(x-1.596))#
