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

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Answer 1 · 2017-01-08T10:58:07

The answer is $=x+1/2ln(∣x+2∣)+5/2ln(∣x-2∣)+C$

Since the degree of the numerator is not less than the degree of the denominator, perform a long division

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

$color(white)(aaaa)$ $x^2$ $color(white)(aaaaaa)$ $-4$ $color(white)(aaaa)$ $∣$ $1$

$color(white)(aaaa)$ $0+3x$ $color(white)(aaa)$ $+4$

Therefore,

By factorising the denominator,

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

Now, we perform the partial fraction decomposition

$(3x+4)/((x+2)(x-2))=A/(x+2)+B/(x-2)$

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

So,

$3x+4=A(x-2)+B(x+2)$

Let $x=2$ , $=>$ , $10=4B$ , $=>$ , $B=5/2$

Let $x=-2$ , $=>$ , $-2=-4A$ , $=>$ , $A=1/2$

Therefore,

$(x^2+3x)/(x^2-4)=1+(1/2)/(x+2)+(5/2)/(x-2)$

So,

$int((x^2+3x)dx)/(x^2-4)=int1dx+(1/2)intdx/(x+2)+(5/2)intdx/(x-2)$

$=x+1/2ln(∣x+2∣)+5/2ln(∣x-2∣)+C$

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