@Jinay Patel
–
no it can't be written as such because if you take 1+x^2 as a variable then you should also manage the dt factor.
–
Gopal Chpidhary
·
3 years, 8 months ago

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@Gopal Chpidhary
–
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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Trevor B.
·
3 years, 8 months ago

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@Trevor B.
–
Actually, the error that Gopal made was with his substitution.

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

\[ \int \frac{1}{1+t} \, dt = \ln (1+t) + C = \ln (1+x^2) + C \]
–
Calvin Lin
Staff
·
3 years, 8 months ago

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TopNewestCan anyone help : https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409 – 柯 南 · 1 year, 11 months ago

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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 – Gopal Chpidhary · 3 years, 8 months ago

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– Calvin Lin Staff · 3 years, 8 months ago

Are you sure? Can you explain your thinking step by step?Log in to reply

– Jinay Patel · 3 years, 8 months ago

ThanksLog in to reply

– Jinay Patel · 3 years, 8 months ago

It can also be written as- log(1+x^2) + cLog in to reply

– Gopal Chpidhary · 3 years, 8 months ago

no it can't be written as such because if you take 1+x^2 as a variable then you should also manage the dt factor.Log in to reply

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. \[\int\dfrac{2x}{x^2+1}dx=\ln|x^2+1|+C\] – Trevor B. · 3 years, 8 months ago

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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

\[ \int \frac{1}{1+t} \, dt = \ln (1+t) + C = \ln (1+x^2) + C \] – Calvin Lin Staff · 3 years, 8 months ago

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