How do you integrate $\int \frac{1 - x^{2}}{(x + 1) (x - 5) (x + 2)}$ $int (1-x^2)/((x+1)(x-5)(x+2))$ using partial fractions?

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Answer 1 · 2016-12-27T08:08:14

The answer is $= - \frac{4}{7} ln (∣ x - 5 ∣) - \frac{3}{7} ln (∣ x + 2 ∣) + C$ $=-4/7ln(∣x-5∣)-3/7ln(∣x+2∣)+ C$

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

Let's rewrite the expression

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

Now we can do the decomposition into partial fractions

$(1-x)/((x-5)(x+2))=A/(x-5)+B/(x+2)$

$=(A(x+2)+B(x-5))/((x-5)(x+2))$

Therefore,

$1-x=A(x+2)+B(x-5)$

Let $x=-2$ , $=>$ , $3=-7B$ , $=>$ , $B=-3/7$

Let $x=5$ , $=>$ , $-4=7A$ , $=>$ , $A=-4/7$

So,

$(1-x)/((x-5)(x+2))=(-4/7)/(x-5)+(-3/7)/(x+2)$

Therefore,

$int((1-x)dx)/((x-5)(x+2))=int((-4/7)dx)/(x-5)+int((-3/7)dx)/(x+2$

$=-4/7ln(∣x-5∣)-3/7ln(∣x+2∣)+ C$

How do you integrate int (1-x^2)/((x+1)(x-5)(x+2)) ∫1−x2(x+1)(x−5)(x+2)int (1-x^2)/((x+1)(x-5)(x+2)) using partial fractions?