Definite Integral Challenge

Method 1:

For the integrand ${\frac{{\left( {16 - {x^2}} \right)x}}{{16 - {x^2} + \sqrt {\left( {4 - x} \right)\left( {4 + x} \right)\left( {12 + {x^2}} \right)} }}}$ , rationalize: $\int_0^2 {\frac{{x\left( {{x^2} - 16} \right)\left( {{x^2} + \sqrt { - {x^4} + 4{x^2} + 192} - 16} \right)}}{{2{x^4} - 36{x^2} + 64}}} \;dx$

Expand the integrand $\int_0^2 {\frac{{x\left( {{x^2} - 16} \right)\left( {{x^2} + \sqrt { - {x^4} + 4{x^2} + 192} - 16} \right)}}{{2{x^4} - 36{x^2} + 64}}}$ : $\int_{}^{} {\left( { - \frac{{16x\sqrt { - {x^4} + 4{x^2} + 192} }}{{2{x^4} - 36{x^2} + 64}} + \frac{{256x}}{{2{x^4} - 36{x^2} + 64}} + \frac{{{x^5}}}{{2{x^4} - 36{x^2} + 64}} + \frac{{{x^3}\sqrt { - {x^4} + 4{x^2} + 192} }}{{2{x^4} - 36{x^2} + 64}} - \frac{{32{x^3}}}{{2{x^4} - 36{x^2} + 64}}} \right)} \;dx$

The anti-derivative of ${ - \frac{{16x\sqrt { - {x^4} + 4{x^2} + 192} }}{{2{x^4} - 36{x^2} + 64}} + \frac{{256x}}{{2{x^4} - 36{x^2} + 64}} + \frac{{{x^5}}}{{2{x^4} - 36{x^2} + 64}} + \frac{{{x^3}\sqrt { - {x^4} + 4{x^2} + 192} }}{{2{x^4} - 36{x^2} + 64}} - \frac{{32{x^3}}}{{2{x^4} - 36{x^2} + 64}}}$ is $\frac{1}{4}\left( {{x^2} + \sqrt { - {x^4} + 4{x^2} + 192} - 14\ln \left( {\sqrt { - {x^4} + 4{x^2} + 192} + 14} \right)} \right) + C$ : $\left. {\left[ {\frac{1}{4}\left( {{x^2} + \sqrt { - {x^4} + 4{x^2} + 192} - 14\ln \left( {\sqrt { - {x^4} + 4{x^2} + 192} + 14} \right)} \right)} \right]} \right|_0^2$

Evaluate the antiderivative! $\left[ {\frac{1}{4}\left( {{2^2} + \sqrt { - {2^4} + 4{{\left( 2 \right)}^2} + 192} - 14\ln \left( {\sqrt { - {{\left( 2 \right)}^4} + 4{{\left( 2 \right)}^2} + 192} + 14} \right)} \right)} \right] - \left[ {\frac{1}{4}\left( {{0^2} + \sqrt { - {0^4} + 4{{\left( 0 \right)}^2} + 192} - 14\ln \left( {\sqrt { - {{\left( 0 \right)}^4} + 4{{\left( 0 \right)}^2} + 192} + 14} \right)} \right)} \right]$ $\left[ {\frac{1}{4}\left( {4 + 8\sqrt 3 - 14\ln \left( {14 + 8\sqrt 3 } \right)} \right)} \right] - \left[ {\frac{1}{4}\left( {8\sqrt 3 - 14\ln \left( {14 + 8\sqrt 3 } \right)} \right)} \right]$ $= \boxed1$

Method 2: Let $I$ denote the given integral. First we make the substitution $y = {x^2}$ , so $dy = 2x\;dx$ . Then: $2I = \int_0^4 {\frac{{16 - y}}{{16 - y + \sqrt {\left( {16 - y} \right)\left( {12 + y} \right)} }}} \;dy = \int_0^4 {\frac{{\sqrt {16 - y} }}{{\sqrt {16 - y} + \sqrt {12 + y} }}\;dy}$

Now make the substitution $z = 4-y$ , so $dz=-dy$ . Then: $2I = \int_0^4 {\frac{{\sqrt {12 + z} }}{{\sqrt {12 + z} + \sqrt {16 - z} }}} \;dz$

Adding the last two equations, we obtain $4I = \int_0^4 {dz} = 4$ and hence $I = \boxed{1}.$