How do you integrate $\frac{2 x^{3} - 3 x^{2} + x + 1}{- 2 x + 1}$ $(2x^3-3x^2+x+1)/(-2x+1)$ ?

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How do you integrate $\frac{2 x^{3} - 3 x^{2} + x + 1}{- 2 x + 1}$ $(2x^3-3x^2+x+1)/(-2x+1)$ ?

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Answer 1 · 2015-09-14T09:49:23

$- \frac{x^{3}}{3} + \frac{x^{2}}{2} - \frac{1}{2} ln | 2 x - 1 | + C$ $-x^3/3+x^2/2-1/2lnabs(2x-1)+C$

Explanation:

We have to divide polinomials:

$(2 x^{3} - 3 x^{2} + x + 1) : (- 2 x + 1) = - x^{2} + x$ $(2x^3-3x^2+x+1):(-2x+1)=-x^2+x$
$- 2 x^{3} + x^{2}$ $-2x^3+x^2$
$- - - -$ $----$
$- 2 x^{2} + x + 1$ $-2x^2+x+1$
$+ 2 x^{2} - x$ $+2x^2-x$
$- - - -$ $----$
$1$ $1$

$\frac{2 x^{3} - 3 x^{2} + x + 1}{- 2 x + 1} = - x^{2} + x + \frac{1}{- 2 x + 1}$ $(2x^3-3x^2+x+1)/(-2x+1)=-x^2+x+1/(-2x+1)$
$\int \frac{2 x^{3} - 3 x^{2} + x + 1}{- 2 x + 1} d x =$ $int(2x^3-3x^2+x+1)/(-2x+1)dx=$
$= \int (- x^{2} + x + \frac{1}{- 2 x + 1}) d x =$ $=int(-x^2+x+1/(-2x+1))dx=$
$= - \int x^{2} d x + \int x d x - \int \frac{d x}{2 x - 1} = I$ $=-intx^2dx+intxdx-intdx/(2x-1)=I$
$2 x - 1 = t, 2 d x = d t, d x = \frac{d t}{2}$ $2x-1=t, 2dx=dt, dx=dt/2$
$I = - \frac{x^{3}}{3} + \frac{x^{2}}{2} - \int \frac{d t}{2} \frac{1}{t} =$ $I=-x^3/3+x^2/2-intdt/2 1/t=$
$= - \frac{x^{3}}{3} + \frac{x^{2}}{2} - \frac{1}{2} ln | 2 x - 1 | + C$ $=-x^3/3+x^2/2-1/2lnabs(2x-1)+C$

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How do you integrate $\frac{2 x^{3} - 3 x^{2} + x + 1}{- 2 x + 1}$ $(2x^3-3x^2+x+1)/(-2x+1)$ ?

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How do you integrate (2x^3-3x^2+x+1)/(-2x+1)2x3−3x2+x+1−2x+1(2x^3-3x^2+x+1)/(-2x+1)?

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How do you integrate $\frac{2 x^{3} - 3 x^{2} + x + 1}{- 2 x + 1}$ $(2x^3-3x^2+x+1)/(-2x+1)$ ?