How do you integrate $(7x-2)/((x-3)^2(x+1))$ using partial fractions?

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Answer 1 · 2017-11-03T17:38:20

Write the partial fractions equation:

$(7x-2)/((x-3)^2(x+1)) = A/(x+1)+B/(x-3)+C/(x-3)^2$

Multiply both sides by $(x-3)^2(x+1)$ :

$7x-2 = A(x-3)^2+B(x-3)(x+1)+C(x+1)$

Let $x = -1$

$7(-1)-2 = A(-1-3)^2+B(-1-3)(-1+1)+C(-1+1)$

$-9 = 16A$

$A = -9/16$

$7x-2 = -9/16(x-3)^2+B(x-3)(x+1)+C(x+1)$

Let $x = 3$

$7(3)-2 = -9/16(3-3)^2+B(3-3)(3+1)+C(3+1)$

#C = 19/4

$7x-2 = -9/16(x-3)^2+B(x-3)(x+1)+19/4(x+1)$

Let $x = 0$

$7(0)-2 = -9/16(0-3)^2+B(0-3)(0+1)+19/4(0+1)$

$-2 = -81/16-3B+19/4(0+1)$

$B = 9/16$

The partial fractions are:

$(7x-2)/((x-3)^2(x+1)) = -9/16 1/(x+1)+9/16 1/(x-3)+19/4 1/(x-3)^2$

Integrate:

$int (7x-2)/((x-3)^2(x+1)) dx = -9/16 int 1/(x+1) dx+9/16 int 1/(x-3) dx +19/4 int 1/(x-3)^2 dx$

The integrals are trivial:

$int (7x-2)/((x-3)^2(x+1)) dx = -9/16 lnt|x+1|+9/16 ln|x-3| -19/4 1/(x-3)+ C$

How do you integrate (7x-2)/((x-3)^2(x+1))(7x-2)/((x-3)^2(x+1)) using partial fractions?