Mathematics

$$\displaystyle \int{\dfrac{(x^{2}-1)dx}{(x^{2}+1)\sqrt{x^{4}+1}}}$$ is equal to :


ANSWER

$$None\ of\ these$$


SOLUTION

We have,

$$\int{\dfrac{\left( {{x}^{2}}-1 \right)dx}{\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{4}}+1}}}$$

Now,

Let

$$ t=\dfrac{x}{\sqrt{{{x}^{4}}+1}} $$

$$ dt=\dfrac{1-{{x}^{4}}}{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{2}}}}dx $$

So,

$$ \int{\dfrac{{{x}^{2}}-1}{\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{4}}+1}}dx=\int{\dfrac{\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)\left( {{x}^{4}}+1 \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{2}}}}}dx} $$

$$ =\int{\dfrac{1}{1+2{{t}^{2}}}.\dfrac{{{x}^{4}}-1}{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{2}}}}}dx $$

$$ =-\int{\dfrac{dt}{1+2{{t}^{2}}}} $$

On integration and we get,

$$ -\int{\dfrac{dt}{1+2{{t}^{2}}}}=-\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\sqrt{2}t $$

$$ =\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{x\sqrt{2}}{\sqrt{{{x}^{4}}+1}}+C $$

Hence, this is the answer.

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