What is #int (-2x^3-x ) / (-4x^2+2x +7 )#?

Question

What is #int (-2x^3-x     ) / (-4x^2+2x  +7     )#?

Luke Alvarez · Accepted Answer

#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))# To find: #int(-2x^3-x)/(-4x^2+2x+7)dx# Let #I=int(-2x^3-x)/(-4x^2+2x+7)dx# dividing the integrand by the fraction 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# rearranging the numerator after taking a negative sign #I_2=int(x^2+x+3.5)/(4x^2-2x-7)dx# Currently, a combination can be used to express the numerator. #(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# Making It Simpler #x^2+x+3.5=4px^2-2px-7p+8qx-2q+r# assembling the terms with similar powers of x #x^2+x+3.5=4px^2+(-2p+8q)x+(-7p-2q+r)# calculating the like powers of x coefficients #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 to make things simpler #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 denominator's squares #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))#

William Abdalla · Answer

undefined The expression $\frac{-2x^3 - x}{-4x^2 + 2x + 7}$ simplifies to $\frac{x(2x^2 + 1)}{(2x - 1)(-2x - 7)}$.