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How do you integrate #x/(x^2+1)#?

Answer 1

Samantha Adkison

#int(x/(x^2+1))dx=1/2ln(x^2+1)+C#

#int(x/(x^2+1))dx#

now #d/(dx)(x^2+1)=2x#

so using #int(f'(x))/(f(x))=ln|f(x)|#

we have #int(x/(x^2+1))dx=1/2ln(x^2+1)+C#

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Answer 2

Axel Alvarez

To integrate ( \frac{x}{x^2 + 1} ), you can use the substitution method. Let ( u = x^2 + 1 ), then ( du = 2x dx ). Rewrite the integral in terms of ( u ) and ( du ). This yields:

[ \int \frac{1}{2} \frac{du}{u} = \frac{1}{2} \ln|u| + C ]

Substitute back ( u = x^2 + 1 ) to get the final result:

[ \frac{1}{2} \ln|x^2 + 1| + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.