Get the answer to "Evaluate ∫4x(x2−1)(x−1)dx"

Pasegeabe85xy

Answered question

2022-04-05

Evaluate

\int \frac{4 x}{(x^{2} - 1) (x - 1)} d x

Answer & Explanation

memantangti17

Beginner2022-04-06Added 13 answers

Another way of splitting the integrals. Write

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

then you have

\int \frac{4 x}{(x^{2} - 1) \cdot (x - 1)} d x = \int \frac{{(x + 1)}^{2}}{(x^{2} - 1) \cdot (x - 1)} d x - \int \frac{{(x - 1)}^{2}}{(x^{2} - 1) \cdot (x - 1)} d x

= \int \frac{x + 1}{{(x - 1)}^{2}} d x - \int \frac{1}{x + 1} d x

= \int \frac{1}{x - 1} d x + 2 \int \frac{1}{{(x - 1)}^{2}} d x - \int \frac{1}{x + 1} d x

Jaslyn Allison

Beginner2022-04-07Added 13 answers

Your initial setup is incorrect, as there can never be constants A,B,C for which

\frac{4 x}{(x^{2} - 1) (x - 1)} = \frac{A x + B}{x^{2} - 1} + \frac{C}{x - 1}

This you can see by multiplying by

x^{2} - 1

. You will see that the right side becomes a polynomial, while the left side does not.
The standard techniques of partial fractions try to get the denominator in the form

{(x - a_{1})}^{r_{1}} {(x - a_{2})}^{r_{2}} \dots {(x - a_{n})}^{r_{n}}

with the

a_{i}

being distinct: this is crucial.
So in your case,

(x^{2} - 1) (x - 1)

becomes

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

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