Integral

**Jake8054** · February 9th 2009, 04:23 PM

(sorry I am not that good with latex)

How do I solve the integral, dx/(1+x^3+sqrt(1+x^6)), from -1 to 1?

**Danny** · February 9th 2009, 05:12 PM

Originally Posted by Jake8054

(sorry I am not that good with latex)

How do I solve the integral, dx/(1+x^3+sqrt(1+x^6)), from -1 to 1?

If the integral is

$\int_{-1}^1 \frac{dx}{1+x^3 + \sqrt{1+x^6}}$

then first rationalize the denominator

$\frac{1}{2} \int_{-1}^1 \frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx$

Then split the integral

$\frac{1}{2} \left( \int_{-1}^0 \frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)$

Then make the substitution $x = -u$ in the first integral

$\frac{1}{2} \left(- \int_{1}^0 \frac{ 1 - u^3 - \sqrt{1+u^6} }{- u^3 }\, du + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)$

or

$\frac{1}{2} \left( -\int_{0}^1 \frac{ 1 - u^3 - \sqrt{1+u^6} }{ u^3 }\, du + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)$

since u is a dummy variable then

$\frac{1}{2} \left( -\int_{0}^1 \frac{ 1 - x^3 - \sqrt{1+x^6} }{ x^3 }\, dx + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)$

so

$\frac{1}{2} \int_{0}^1 \frac{ 2x^3 }{ x^3 }\, dx = 1$

**Jake8054** · February 9th 2009, 05:18 PM

zomg i cant believe it was that simple. Thank you so much!

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