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))
