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\int \frac{x^{4}dx}{x^{4}-1}

  • Option 1)

    x+\frac{1}{2}\tan ^{-1}x+\frac{1}{4}\ln \left ( x+1 \right )-\frac{1}{4}\ln \left ( x-1 \right )+C

  • Option 2)

    x-\frac{1}{2}\tan ^{-1}x-\frac{1}{4}\ln \left ( x+1 \right )+\frac{1}{4}\ln \left ( x-1 \right )+C

  • Option 3)

    x-\frac{1}{2}\tan ^{-1}x+\frac{1}{4}\ln \frac{\left ( x+1 \right )}{x-1}+C

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learned,

 

Rule for Partial fraction -

Quadratic and Non-repeated.

Let Q(x)=(x-\alpha _{1})(x-\alpha _{2})\cdot \cdot \cdot \cdot (ax^{2}+bx+c)

Then,

\frac{P(x)}{Q(x)}=\frac{Ax+B}{ax^{2}+bx+c}+\frac{c}{x-\alpha _{1}}+\frac{D}{x-\alpha _{2}}+\cdot \cdot \cdot

- wherein

Find A,B,C\cdot \cdot \cdot

By comparing with P(x)

 

 

\int \frac{x^{4}-1dx}{x^{4}-1}+\int \frac{dx}{x^{4}-1}=x+\int \frac{dx}{\left ( x-1 \right )\left ( x+1 \right )\left ( x^{2}+1 \right )}

\frac{Ax+B}{x^{2}+1}+\frac{C}{x+1}+\frac{D}{x-1}=\frac{1}{\left ( x-1 \right )\left ( x+1 \right )\left ( x^{2}+1 \right )}

On solving A=0,\: B=\frac{-1}{2},\: C=\frac{-1}{4},\: D=\frac{1}{4}

Thus I=x-\frac{1}{2}\int \frac{dx}{x^{2}+1}-\frac{1}{4}\int \frac{dx}{\left ( x+1 \right )}+\frac{1}{4}\int \frac{dx}{\left ( x-1 \right )}

I=x-\frac{1}{2}\tan ^{-1}x-\frac{1}{4}\ln \left ( x+1 \right )+\frac{1}{4}\ln \left ( n-1 \right )+C


Option 1)

x+\frac{1}{2}\tan ^{-1}x+\frac{1}{4}\ln \left ( x+1 \right )-\frac{1}{4}\ln \left ( x-1 \right )+C

Option 2)

x-\frac{1}{2}\tan ^{-1}x-\frac{1}{4}\ln \left ( x+1 \right )+\frac{1}{4}\ln \left ( x-1 \right )+C

Option 3)

x-\frac{1}{2}\tan ^{-1}x+\frac{1}{4}\ln \frac{\left ( x+1 \right )}{x-1}+C

Option 4)

none of these

Posted by

gaurav

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