How do you integrate $int (4x^3 + 12x^2 - x - 4 )/( 4x^2 - 1)$ using partial fractions?

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Answer 1 · 2016-12-12T06:47:02

The answer is $=x^2/2+3x+1/4ln(∣2x+1∣)-1/4ln(∣2x-1∣)+C$

We use $a^2-b^2=(a+b)(a-b)$

$intx^ndx=x^(n+1)/(n+1)$ and $intdx/x=∣x∣$

Let's do a long division

$color(white)(aaaa)$ $4x^3-x+12x^2-4$ $color(white)(aaaa)$ $∣$ $4x^2-1$

$color(white)(aaaa)$ $4x^3-x$ $color(white)(aaaaaaaaaaaaaa)$ $∣$ $x+3$

$color(white)(aaaaaa)$ $0 -0+12x^2-4$

$color(white)(aaaaaaaaaaa)$ $+12x^2-3$

$color(white)(aaaaaaaaaaaaaa)$ $+0-1$

So,

$(4x^3-x+12x^2-4)/(4x^2-1)=x+3-1/(4x^2-1)$

$=x+3-1/((2x+1)(2x-1))$

Let's do the decomposition into partial fractions

$1/((2x+1)(2x-1))=A/(2x+1)+B/(2x-1)$

$=(A(2x-1)+B(2x+1))/((2x+1)(2x-1))$

Therefore,

$1=A(2x-1)+B(2x+1)$

Let $x=1/2$ , $=>$ , $1=2B$ , $=>$ , $B=1/2$

Let $x=-1/2$ , $=>$ , $1=-2A$ , $=>$ , $A=-1/2$

So,

$(4x^3-x+12x^2-4)/(4x^2-1)=x+3-((-1/2)/(2x+1)+(1/2)/(2x-1))$

$int((4x^3-x+12x^2-4)dx)/(4x^2-1)=int(x+3)dx+1/2int(dx)/(2x+1)-1/2int(dx)/(2x-1)$

$=x^2/2+3x+1/4ln(∣2x+1∣)-1/4ln(∣2x-1∣)+C$

How do you integrate int (4x^3 + 12x^2 - x - 4 )/( 4x^2 - 1)int (4x^3 + 12x^2 - x - 4 )/( 4x^2 - 1) using partial fractions?