@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.
\[\int\dfrac{2x}{x^2+1}dx=\ln|x^2+1|+C\]

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

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

Log in to reply

Thanks

Log in to reply

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

Log in to reply

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. \[\int\dfrac{2x}{x^2+1}dx=\ln|x^2+1|+C\]

Log in to reply

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

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

Log in to reply