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How do you integrate #int4x(x^2+3)^(-3) dx#?

Answer 1

Emma Aldridge

#-1/(x^2+3)^2 + C#

#int4x(x^2+3)^(-3) dx = 2 int (x^2+3)^(-3) 2xdx =#

#= 2 int (x^2+3)^(-3) (x^2+3)'dx = 2 int (x^2+3)^(-3) d(x^2+3) =#

#= 2 (x^2+3)^-2/-2 +C = -1/(x^2+3)^2 + C#

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Answer 2

Adam Adkins

To integrate ( \int 4x(x^2+3)^{-3} , dx ), perform the substitution ( u = x^2 + 3 ). Then, ( du = 2x , dx ). Rewrite the integral in terms of ( u ). The integral becomes ( \int \frac{2}{u^3} , du ). Integrate ( \frac{2}{u^3} ) with respect to ( u ) to get ( -\frac{1}{u^2} ). Finally, substitute back ( x^2 + 3 ) for ( u ) to get the final answer: ( -\frac{1}{x^2 + 3} + C ).

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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.