How do you integrate #(xdx)/(x^2+2)^3# which has the upper and lower limits, 2 and 1 ?

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Answer 1 · 2018-04-03T01:42:51

The indefinite integral is equal to #1/48#.

First. compute the indefinite intregral:

#color(white)=intx/(x^2+2)^3dx#

Let #u=x^2+2#, so #du=2xdx# or #dx=1/(2x)du#:

#=intx/u^3 1/(2x)du#

#=intcolor(red)cancelcolor(black)x/u^3 1/(2color(red)cancelcolor(black)x)du#

#=1/2int1/u^3du#

#=1/2intu^-3du#

#=1/2*(u^(-3+1))/(-3+1)#

#=1/2*(u^(-2))/(-2)#

#=-u^-2/4#

#=-1/(4u^2)#

Plug in #x^2+2# back in for #u#, and don't forget to add #C#:

#=-1/(4(x^2+2)^2)+C#

Now, plug in #x=2# to this and then subtract the value when you plug in #x=1#:

#=-1/(4(2^2+2)^2)-(-1/(4(1^2+2)^2))#

#=-1/(4(4+2)^2)+1/(4(1+2)^2)#

#=-1/(4*6^2)+1/(4*3^2)#

#=-1/(4*36)+1/(4*9)#

#=-1/144+1/36#

#=-1/144+4/144#

#=3/144#

#=1/48#

That's the result. Hope this helped!