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 $n!=-1$ )

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

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

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

$=1/6$

What is int_0^1 (1+3(x^2)) ^ -(3/2)x dxint_0^1 (1+3(x^2)) ^ -(3/2)x dx?