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How do you find the integral #ln(x^2+4)#?

Answer 1

Cameron Alexander

#I = xln(x^2+4)- 2x + 4arctan(x/2)+C#

#I=int ln(x^2+4)dx#

#u=ln(x^2+4) => du=1/(x^2+4)*2x*dx= (2x)/(x^2+4)dx#

#dv=dx => v=x#

#I=xln(x^2+4)- int x(2x)/(x^2+4)dx#

#I = xln(x^2+4)- 2int x^2/(x^2+4)dx#

#I = xln(x^2+4)- 2int (x^2+4-4)/(x^2+4)dx#

#I = xln(x^2+4)- 2int (1-4/(x^2+4))dx#

#I = xln(x^2+4)- 2(x-4*1/2arctan(x/2))+C#

#I = xln(x^2+4)- 2x + 4arctan(x/2)+C#

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Answer 2

Scarlett Allison

To find the integral of ln(x^2 + 4), you can use integration by parts. Let u = ln(x^2 + 4) and dv = dx. Then differentiate u to get du and integrate dv to get v. After that, apply the integration by parts formula: ∫u dv = uv - ∫v du. Substituting the values, you'll get the integral of ln(x^2 + 4).

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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.