# Thread: Integral

1. ## Integral

(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?

2. 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

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

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