@Gopal Chpidhary
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Jinay had the correct answer. Gopal didn't account for the \(2x\) in the numerator. What he integrated was this. Note the \(1\) in the numerator instead of \(2x\)
\[\int\dfrac{1}{x^2+1}dx=\arctan x+C\]

Here is the solution to your problem.

Let \(u=x^2+1\). Thus, \(du=2xdx\). The integral becomes \[\int\dfrac{du}{u}\] which equals \(\ln|u|+C\). Substituting \(u=x^2+1\) gives the following.
\[\int\dfrac{2x}{x^2+1}dx=\ln|x^2+1|+C\]

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TopNewestCan anyone help : https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409

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its simple take x^2 = t then, 2xdx=dt now integrate 1\2(1+t^2) it will become tan-1x^\2 + c

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Are you sure? Can you explain your thinking step by step?

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Thanks

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It can also be written as- log(1+x^2) + c

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Here is the solution to your problem.

Let \(u=x^2+1\). Thus, \(du=2xdx\). The integral becomes \[\int\dfrac{du}{u}\] which equals \(\ln|u|+C\). Substituting \(u=x^2+1\) gives the following. \[\int\dfrac{2x}{x^2+1}dx=\ln|x^2+1|+C\]

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He forgot that \(x^2 = t \) in the denominator, which should have become \( 1+t \) instead of \( 1 + t^2 \).

Once that is fixed, we are looking for

\[ \int \frac{1}{1+t} \, dt = \ln (1+t) + C = \ln (1+x^2) + C \]

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