How do you find the antiderivative of #int 1/(x^2+10x+25) dx#?

Answer 1

#-1/(x+5)+C#

Notice the denominator can be factored:

#int1/(x^2+10x+25)dx=int1/(x+5)^2dx=int(x+5)^-2dx#
Now we can use substitution. Let #u=x+5#. Thus, #du=dx#. Substituting them in:
#=intu^-2du#
Now use the power rule for integration #intu^ndu=u^(n+1)/(n+1)+C#
#=u^(-2+1)/(-2+1)+C=u^(-1)/(-1)+C=-1/u+C=-1/(x+5)+C#
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Answer 2

To find the antiderivative of (\int \frac{1}{x^2+10x+25} , dx), you can use the method of partial fraction decomposition. First, factor the denominator as ((x+5)^2). Then, express the fraction as the sum of two simpler fractions, (\frac{A}{x+5}) and (\frac{B}{(x+5)^2}). Solving for (A) and (B) by equating coefficients, you'll find that (A = -\frac{1}{10}) and (B = \frac{1}{10}). So the integral becomes (\int \frac{-1/10}{x+5} + \frac{1/10}{(x+5)^2} , dx). Finally, integrate each term separately to get the antiderivative, which is (-\frac{1}{10} \ln |x+5| - \frac{1}{10(x+5)} + C), where (C) is the constant of integration.

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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.

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.

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.

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.

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