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

Question

Ezekiel Alexander · Accepted Answer

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# Power rule: #=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!

Emma Aldridge · Answer

undefined To integrate the function $\frac{xdx}{(x^2+2)^3}$ with limits from 1 to 2, you can use a trigonometric substitution. Let $x = \sqrt{2}\tan(\theta)$. Then $dx = \sqrt{2}\sec^2(\theta) d\theta$. Substitute these expressions into the integral and simplify. After integration, apply the limits of integration from 1 to 2.