Question

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Answer 1

Question :

Evaluate :

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

Solution :

We have , to integrate :

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

$\sf\dfrac{x^3+x+1}{x^2-1}dx$

By long division

$\begin{array}{c | c c c c c c c} \cline{2-8}&x+&&&&&& \\ \cline{1-8} (x^2 +1) & x^3&+x&+1&&&&\\ & x^3& -x \\ &(-)&(+) \\ \cline{2-8} &&2x&+1\\ \\ \cline{2-8} \end{array}$

Dividend = Divisor × Quotient + Remainder

$\sf\:x^3+x+1=x^2-1\:\times\:x+(2x+1)$

Divide both sides by $\sf\:x^2-1$

$\sf\:\dfrac{x^3+x+1}{x^2-1}=x+\dfrac{2x+1}{x^2-1}$

Thus ,

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

$\sf\int\:(x+\dfrac{2x+1}{x^2-1})dx$

$\sf\int\:x\:dx+\int\dfrac{2x+1}{x^2-1}dx$

$\sf\int\:x\:dx+\int\dfrac{x+1+x}{x^2-1}dx$

$\sf\int\:x\:dx+\int\dfrac{x+1}{x^2-1}dx+\dfrac{x}{x^2-1}dx$

$\sf\:\green{Now\:Solving}$

$\sf\int\dfrac{x+1}{x^2-1}dx$

$\sf\int\dfrac{x+1}{(x+1)(x-1)}dx$

$\sf\int\dfrac{1}{x-1}dx$

$\sf=\log(x-1)$

$\sf\blue{Now\:Solving}$

$\sf\int\dfrac{x}{x^2-1}dx$

$\sf\dfrac{1}{2}\int\dfrac{2x}{x^2-1}dx$

Let t=x²-1

thus, dt/dx=2x

$\sf\dfrac{1}{2}\int\dfrac{2x}{t}\times\dfrac{dt}{2x}$

$\sf\dfrac{1}{2}\int\dfrac{1}{t}dt$

$\sf=\dfrac{\log\:t}{2}$

$\sf=\dfrac{\log(x^2-1)}{2}$

$\sf=\dfrac{\log(x+1)}{2}+\dfrac{\log(x-1)}{2}$

Hence,

$\sf\int\:x\:dx+\int\dfrac{x+1}{x^2-1}dx+\dfrac{x}{x^2-1}dx$

$\sf=\dfrac{x^2}{2}+\log(x-1)+\dfrac{\log(x+1)}{2}+\dfrac{\log(x-1)}{2}+c$

$\sf=\dfrac{x^2+3\log(x-1)+\log(x+1)}{2}+c$

Therefore,

$\tt\dfrac{x^3+x+1}{x^2-1}dx$

$\sf=\dfrac{x^2+3\log(x-1)+\log(x+1)}{2}+c$

Answer 2

The answer after evaluating will be :

x²+3log(x-1)+log(x+1)+c÷2

Go through the attachment for the answer.