What is $\int_{0}^{1} {(1 + 3 (x^{2}))}^{- (\frac{3}{2})} x d x$ $int_0^1 (1+3(x^2)) ^ -(3/2)x dx$ ?

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What is $\int_{0}^{1} {(1 + 3 (x^{2}))}^{- (\frac{3}{2})} x d x$ $int_0^1 (1+3(x^2)) ^ -(3/2)x dx$ ?

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Answer 1 · 2016-02-05T07:04:07

$\int_{0}^{1} {(1 + 3 x^{2})}^{- \frac{3}{2}} x d x = \frac{1}{6}$ $int_0^1(1+3x^2)^(-3/2)xdx = 1/6$

Explanation:

We will proceed by using substitution.

Let $u = 1 + 3 x^{2}$ $u = 1+3x^2$
Then $d u = 6 x d x \Rightarrow x d x = \frac{1}{6} d u$ $du = 6xdx => xdx = 1/6du$
At $x = 0$ $x = 0$ we have $u = 1$ $u = 1$
At $x = 1$ $x = 1$ we have $u = 4$ $u = 4$

Then, performing the substitution, we have

$\int_{0}^{1} {(1 + 3 x^{2})}^{- \frac{3}{2}} x d x = \int_{1}^{4} \frac{u^{- \frac{3}{2}}}{6} d u$ $int_0^1(1+3x^2)^(-3/2)xdx = int_1^4u^(-3/2)/6du$

$= \frac{1}{6} \int_{1}^{4} u^{- \frac{3}{2}} d u$ $=1/6int_1^4u^(-3/2)du$

$= \frac{1}{6} {[\frac{u^{- \frac{1}{2}}}{- \frac{1}{2}}]}_{1}^{4}$ $=1/6[u^(-1/2)/(-1/2)]_1^4$ (as $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n+1)/(n+1)+C$ for $n \neq - 1$ $n!=-1$ )

$= - \frac{1}{3} (4^{- \frac{1}{2}} - 1^{- \frac{1}{2}})$ $=-1/3(4^(-1/2)-1^(-1/2))$

$= - \frac{1}{3} (\frac{1}{2} - 1)$ $=-1/3(1/2 - 1)$

$= - \frac{1}{3} (- \frac{1}{2})$ $=-1/3(-1/2)$

$= \frac{1}{6}$ $=1/6$

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What is $\int_{0}^{1} {(1 + 3 (x^{2}))}^{- (\frac{3}{2})} x d x$ $int_0^1 (1+3(x^2)) ^ -(3/2)x dx$ ?

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Explanation:

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What is $\int_{0}^{1} {(1 + 3 (x^{2}))}^{- (\frac{3}{2})} x d x$ $int_0^1 (1+3(x^2)) ^ -(3/2)x dx$ ?