How do you integrate $\int \frac{d x}{{(a x^{2} + x^{2})}^{\frac{3}{2}}}$ $int dx/(ax^2+x^2)^(3/2)$ using trig substitutions?

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How do you integrate $\int \frac{d x}{{(a x^{2} + x^{2})}^{\frac{3}{2}}}$ $int dx/(ax^2+x^2)^(3/2)$ using trig substitutions?

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Answer 1 · 2016-09-12T14:55:26

$\frac{- 1}{2 x^{2} {(a + 1)}^{\frac{3}{2}}} + C$ $(-1)/(2x^2(a+1)^(3/2))+C$

Explanation:

Using a trig substitution would not work here. However, simplifying this reveals that there is a simpler, yet sneakier, solution.

$\int \frac{d x}{{(a x^{2} + x^{2})}^{\frac{3}{2}}} = \int \frac{d x}{{(x^{2} (a + 1))}^{\frac{3}{2}}} = \int \frac{d x}{{(x^{2})}^{\frac{3}{2}} {(a + 1)}^{\frac{3}{2}}}$ $intdx/(ax^2+x^2)^(3/2)=intdx/(x^2(a+1))^(3/2)=intdx/((x^2)^(3/2)(a+1)^(3/2))$

Note that ${(a + 1)}^{\frac{3}{2}}$ $(a+1)^(3/2)$ is a constant:

$= \frac{1}{{(a + 1)}^{\frac{3}{2}}} \int \frac{d x}{x^{3}} = \frac{1}{{(a + 1)}^{\frac{3}{2}}} \int x^{- 3} d x$ $=1/(a+1)^(3/2)intdx/x^3=1/(a+1)^(3/2)intx^-3dx$

$= \frac{1}{{(a + 1)}^{\frac{3}{2}}} (\frac{x^{- 2}}{- 2}) = \frac{- 1}{2 x^{2} {(a + 1)}^{\frac{3}{2}}} + C$ $=1/(a+1)^(3/2)((x^-2)/(-2))=(-1)/(2x^2(a+1)^(3/2))+C$

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How do you integrate $\int \frac{d x}{{(a x^{2} + x^{2})}^{\frac{3}{2}}}$ $int dx/(ax^2+x^2)^(3/2)$ using trig substitutions?

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Explanation:

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