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

Question

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

Samantha Adkison · Accepted Answer

#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#

Emily Ammerman · Answer

undefined To evaluate $ \int_{0}^{1} (1+3x^2)^{-\frac{3}{2}}x \, dx $, we can use the substitution method. Let $ u = 1 + 3x^2 $. Then, $ du = 6x \, dx $ or $ \frac{du}{6} = x \, dx $.

When $ x = 0 $, $ u = 1 $, and when $ x = 1 $, $ u = 1 + 3(1)^2 = 4 $.

So, the integral becomes:

$$ \frac{1}{6} \int_{1}^{4} u^{-\frac{3}{2}} \, du $$

Now, we can integrate $ u^{-\frac{3}{2}} $ with respect to $ u $:

$$ \frac{1}{6} \int_{1}^{4} u^{-\frac{3}{2}} \, du = \frac{1}{6} \left[ -\frac{2}{\sqrt{u}} \right]_{1}^{4} $$

$$ = \frac{1}{6} \left( -\frac{2}{\sqrt{4}} + \frac{2}{\sqrt{1}} \right) $$

$$ = \frac{1}{6} \left( -\frac{1}{\sqrt{2}} + 2 \right) $$

$$ = \frac{1}{6} \left( 2 - \frac{1}{\sqrt{2}} \right) $$

$$ = \frac{1}{3} - \frac{1}{6\sqrt{2}} $$