What is #int (2x^3-3x^2-4x-3 ) / (-6x^2+ 3 x -4 )dx#?

Question

What is #int (2x^3-3x^2-4x-3    ) / (-6x^2+  3  x  -4     )dx#?

Elijah Abram · Accepted Answer

#-x^2/6+x/3+19/18ln|(1/2sqrt(29/2))/sqrt(6x^2-3x+4)|-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.#

First we should get in the numerator a polynomial of a grade inferior than the denominator's

By long division:

#" "2x^3-3x^2-4x-3" "#|#" "-6x^2+3x-4# #-2x^3+x^2-4/3x" "#|____ _____#" "-1/3x+1/3# #" "-2x^2-16/3x-3# #" "2x^2-x+4/3# #" "# _______ #" "-19/3x-5/3=-1/3(19x+5)#

So the expression becomes #=1/3int (-x+1)dx-1/3*1/6int(19x+5)/(x^2-x/2+2/3)dx#

Dealing with #=int(19x+5)/(x^2-x/2+2/3)dx# Since #(x-1/4)^2=x^2-x/2+1/16# and #2/3-1/16=29/48=(1/4sqrt(29/3))^2# We can make

#x-1/4=1/4sqrt(29/3)*tan y# => #dx=1/4sqrt(29/3)*sec^2 y*dy# Now let's find how many units of #(x-1/4)# there are in the denominator #" "19x+5" "#|#" "x-1/4# #-19x+19/4" "#|____ _____#" "19# #" "39/4#

So the partial expression becomes #=19int ((1/4sqrt(29/3)tan y)(1/4sec^2 y))/(1/16*29/3sec^2 y)dy+39/4int (1/4sqrt(29/3))/(1/16*29/3sec^2 y)dy# #=-19ln|cos y|+39sqrt(3/29)y# But #tan y = sqrt(3/29)(4x-1)# => #siny =sqrt(3/29)(4x-1)cosy# And #sin^2y+cos^2y=1# => #(3/29(16x^2-8x+1)+1)cos^2 y=1# => #(48x^2-24x+32)/29*cos^2 y=1# => #cos y=(1/2sqrt(29/2))/sqrt(6x^2-4x+8# Then the partial expression becomes #=-19ln|(1/2sqrt(29/2))/sqrt(6x^2-4x+8)+39sqrt(3/29)tan^(-1) (sqrt(3/29)(4x-1))#

Back to the main expression: #=-x^2/6+1/3-1/18[-19ln|(1/2sqrt(29/2))/sqrt(6x^2-4x+8)+39sqrt(3/29)tan^(-1) (sqrt(3/29)(4x-1))]# #=-x^2/6+x/3+19/18ln|(1/2sqrt(29/2))/sqrt(6x^2-3x+4)|-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.#

One step further: simplifying The result above is good enough, but the constant part of the logarithm can be absorbed by the general constant: #=-x^2/6+x/3+19/18[ln(1/2sqrt(29/2))-(1/2)ln|6x^2-3x+4|]-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.#

#=-x^2/6+x/3-19/36ln|6x^2-3x+4|-13/(2sqrt(87))*tan^(-1) (4sqrt(3/29)(x-1/4))+const.#

Christopher Allers · Answer

undefined To find the integral of the given expression $\frac{2x^3 - 3x^2 - 4x - 3}{-6x^2 + 3x - 4}$ with respect to $x$, you can use polynomial long division to simplify the expression first. Once simplified, you can then integrate each term separately.