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The integral value of \int\frac{x\ dx}{\sqrt{x^{4}+1}}

Option: 1

\frac{1}{2}log\left |x^{2}-\sqrt{x^{4}+1} \right |+C


Option: 2

-\frac{1}{2}log\left |x^{2}-\sqrt{x^{4}+1} \right |+C


Option: 3

\frac{1}{2}log\left |x^{2}+\sqrt{x^{4}+1} \right |+C


Option: 4

-\frac{1}{2}log\left |x^{2}+\sqrt{x^{4}+1} \right |+C


Answers (1)

best_answer

Given integral,
\int\frac{x\ dx}{\sqrt{x^{4}+1}}
\int\frac{x\ dx}{\sqrt{x^{4}+1}}=\frac{1}{2}\int\frac{2x\ dx}{\left (\sqrt{\left (x^{2} \right )^{2}+1} \right )}

Using the formula,
\int\frac{dx}{\sqrt{x^{2}+a^{2}}}=log\left |x+{\sqrt{x^{2}+a^{2}}} \right |+C
Therefore,
\int\frac{x\ dx}{\sqrt{x^{4}+1}}=\frac{1}{2}log\left |x^{2}+\sqrt{x^{4}+1} \right |+C

Posted by

SANGALDEEP SINGH

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