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How do you evaluate #int x/(x^2+2)^3 dx# for [-2, -1]?

Answer 1

Samantha Adkison

#-33/3136#

Use substitution technique and let #u=x^2+3# Then #du=2xdx# #therefore 1/2du=xdx#

Limits : #x=-1=>u=4 and x=-2=>u=7#

We may hence re-write this integral as

#1/2int_4^7u^(-3)du=1/2[(u^(-2))/(-2)]_7^4#

#=-1/4(1/16-1/49)=-33/3136#

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Answer 2

Emma Aldridge

To evaluate the integral (\int_{-2}^{-1} \frac{x}{(x^2 + 2)^3} dx), you can use the substitution method. Let (u = x^2 + 2), then (du = 2x , dx). This transforms the integral into (\frac{1}{2} \int \frac{1}{u^3} du). Now, integrate (\frac{1}{u^3}) with respect to (u) and then replace (u) with (x^2 + 2). Finally, evaluate the expression at the upper and lower limits of integration and subtract the result at the lower limit from the result at the upper limit.

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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.