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

Answer 1

Elijah Abram

By parts.

You can integrate it by parts with the rule

#\int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)#

where we assume that

#f'(x)=1# and #g(x)=ln(x+sqrt(x^2+1))#

consequently

#f(x)=x# and #g'(x)=1/(sqrt(x^2+1))#.

The integral is then

#\int ln(x+sqrt(x^2+1))dx#

#=x ln(x+sqrt(x^2+1))-\int x/(sqrt(x^2+1))dx#

#=x ln(x+sqrt(x^2+1))-sqrt(x^2+1) + C#.

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Answer 2

Aurora Alvarado

To integrate ln(x+sqrt(x^2+1)), you can use integration by parts. Let u = ln(x + sqrt(x^2 + 1)) and dv = dx. Then, differentiate u to find du, and integrate dv to find v. After that, use the integration by parts formula:

∫udv = uv - ∫vdu

Finally, substitute the values of u, v, du, and dv into the formula and solve for the integral.

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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.