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How do you evaluate the integral #int 1/(x^2+2x+1)dx# from -1 to 0?

Answer 1

There is a problem as #x!=-1#

Let's factorise the denominator #x^2+2x+1=(x+1)^2# We use #intx^ndx=x^(n+1)/(n+1) (x!=-1)# There is a problem as #x!=-1#

#int_-1^0dx/(x+1)^2=(-1/(x+1))_-1^0=-1+oo#

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Answer 2

Samantha Adkison

To evaluate the integral (\int_{-1}^{0} \frac{1}{x^2 + 2x + 1} , dx) from -1 to 0, first factor the denominator (x^2 + 2x + 1), which becomes ((x + 1)^2). Then, rewrite the integral as (\int_{-1}^{0} \frac{1}{(x + 1)^2} , dx). Now, apply the formula for the integral of (1/(x + a)^n), which is (-1/(n - 1)(x + a)^{n - 1}).

Substituting (a = -1) and (n = 2), the integral becomes (-1/(2 - 1)(x + 1)^{2 - 1}) evaluated from -1 to 0.

Evaluating this from -1 to 0, you get (-1/1(0 + 1)^{1} - (-1/1(-1 + 1)^{1})), which simplifies to (-1/1(1) - (-1/1(1))).

Solving this yields (-1 - (-1)), which equals (0). Thus, the value of the integral is (0).

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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.