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

G gaurav

As we learned,

Rule for Partial fraction -

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

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