Question

integrate the rational functions ~​

Answer 1

Answer:

Hello mate ,

see the above attachment! !

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Answer 2

Answer:

$\large\bold\red{ \frac{ {x}^{2} }{2} + \frac{3}{2} ln |x - 1| + \frac{1}{2} ln |x + 1| + c}$

Step-by-step explanation:

Given,

$\int \frac{ {x}^{3} + x + 1 }{ {x}^{2} - 1} dx$

Further simplifying,

We get,

$= \int \frac{ {x}^{3} - x + 2x + 1}{ {x}^{2} - 1 } dx \\ \\ = \int \frac{x( {x}^{2} - 1) + (2x + 1)}{( {x}^{2} - 1) } dx \\ \\ = \int xdx + \int \frac{2x}{ {x}^{2} - 1 } dx + \int \frac{1}{ {x}^{2} - 1} dx$

Therefore,

Putting the integral values of respective terms,

We get,

$= \frac{ {x}^{2} }{2} + ln | {x}^{2} - 1 | + \frac{1}{2} ln | \frac{x - 1}{x + 1} | + c \\ \\ = \frac{ {x}^{2} }{2} + ln |(x - 1)(x + 1)| + \frac{1}{2} ln | \frac{x - 1}{x + 1} | + c$

But,

We know that,

• $ln( \alpha \beta ) = ln( \alpha ) + ln( \beta )$
• $ln( \frac{ \alpha }{ \beta } ) = ln( \alpha ) - ln( \beta )$

Therefore,

We get,

$= \frac{ {x}^{2} }{2} + ln |x - 1| + ln |x + 1| + \frac{1}{2} |x - 1| - \frac{1}{2} |x + 1| + c \\ \\ = \large\bold{ \frac{ {x}^{2} }{2} + \frac{3}{2} ln |x - 1| + \frac{1}{2} ln |x + 1| + c}$

Where,

C is integral constant