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How do you integrate #x^3/((x^2+5)^2)#?

Answer 1

Christopher Agresta

#int x^3/(x^2+5)^2dx#

We are able to rewrite:

#x^3/(x^2+5)^2= (betax+gamma)/(x^2+5) + (thetax+nu)/(x^2+5)^2#

So : #((betax+gamma)(x^2+5))/((x^2+5)(x^2+5))+(thetax+nu)/(x^2+5)^2#

Then : #(betax^3+gammax^2+(5beta+theta)x+nu+5gamma)/(x^2+5)^2#

Recognition:

#nu + 5gamma = 0# #5beta + theta = 0# #gamma = 0# #beta = 1#

So :

#beta = 1# # gamma = 0# # theta = -5 # # nu = 0#

#int x^3/(x^2+5)^2dx = intx/(x^2+5)dx-5int(x)/(x^2+5)^2dx#

#1/2int (2x)/(x^2+5)dx -5/2int(2x)/(x^2+5)^2dx#

#= (1/2ln(x^2+5)+5/(2x^2+10))+C#

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Answer 2

Adam Adkins

To integrate ( \frac{x^3}{(x^2 + 5)^2} ), you can use the method of substitution. Let ( u = x^2 + 5 ), then ( du = 2x , dx ).

Now, rewrite the integral in terms of ( u ):

[ \int \frac{x^3}{(x^2 + 5)^2} , dx = \frac{1}{2} \int \frac{1}{u^2} , du ]

This becomes a straightforward integral:

[ \frac{1}{2} \int \frac{1}{u^2} , du = -\frac{1}{2u} + C ]

Now, substitute back for ( u ):

[ -\frac{1}{2(x^2 + 5)} + 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.