Integration

Integrate the following expression- 2x/(1+x^2)

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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

柯南 - 5 years, 9 months ago

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 - 7 years, 6 months ago

Thanks

jinay patel - 7 years, 6 months ago

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

jinay patel - 7 years, 6 months ago

@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 - 7 years, 6 months ago

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

Trevor B. - 7 years, 6 months ago

@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 - 7 years, 6 months ago

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

Calvin Lin Staff - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$