What is #int_0^1 (1+3(x^2)) ^ -(3/2)x dx#?

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Answer 1 · 2016-02-05T07:04:07

#int_0^1(1+3x^2)^(-3/2)xdx = 1/6#

We will proceed by using substitution.

Let #u = 1+3x^2#
Then #du = 6xdx => xdx = 1/6du#
At #x = 0# we have #u = 1#
At #x = 1# we have #u = 4#

Then, performing the substitution, we have

#int_0^1(1+3x^2)^(-3/2)xdx = int_1^4u^(-3/2)/6du#

#=1/6int_1^4u^(-3/2)du#

#=1/6[u^(-1/2)/(-1/2)]_1^4# (as #intx^ndx = x^(n+1)/(n+1)+C# for #n!=-1#)

#=-1/3(4^(-1/2)-1^(-1/2))#

#=-1/3(1/2 - 1)#

#=-1/3(-1/2)#

#=1/6#