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# Integration

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

Note by Jinay Patel
3 years, 8 months ago

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Can anyone help : https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409 · 1 year, 11 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 · 3 years, 8 months ago

Are you sure? Can you explain your thinking step by step? Staff · 3 years, 8 months ago

Thanks · 3 years, 8 months ago

It can also be written as- log(1+x^2) + c · 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. · 3 years, 8 months ago

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. $\int\dfrac{2x}{x^2+1}dx=\ln|x^2+1|+C$ · 3 years, 8 months ago

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

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