Answer
$$\frac{1}{{12}}\ln 7$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{1}{{16 - 9{x^2}}}} dx \cr
& \int_{ - 1}^1 {\frac{1}{{16 - 9{x^2}}}} dx = \int_{ - 1}^1 {\frac{1}{{{{\left( 4 \right)}^2} - {{\left( {3x} \right)}^2}}}} dx \cr
& {\text{Let }}u = 3x,{\text{ }}du = 3dx,{\text{ }}du = \frac{1}{3}du \cr
& {\text{The new limits of integration are}} \cr
& x = 1 \to u = 3 \cr
& x = - 1 \to u = - 3 \cr
& {\text{Substituting}} \cr
& \int_{ - 1}^1 {\frac{1}{{{{\left( 4 \right)}^2} - {{\left( {3x} \right)}^2}}}} dx = \int_{ - 3}^3 {\frac{1}{{{{\left( 4 \right)}^2} - {u^2}}}} \left( {\frac{1}{3}} \right)du \cr
& = \frac{1}{3}\int_{ - 3}^3 {\frac{1}{{{{\left( 4 \right)}^2} - {u^2}}}} du \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{{a^2} - {u^2}}} = } \frac{1}{{2a}}\ln \left| {\frac{{a + u}}{{a - u}}} \right| + C \cr
& \frac{1}{3}\int_{ - 3}^3 {\frac{1}{{{{\left( 4 \right)}^2} - {u^2}}}} du = \frac{1}{3}\left[ {\frac{1}{{2\left( 4 \right)}}\ln \left| {\frac{{4 + u}}{{4 - u}}} \right|} \right]_{ - 3}^3 \cr
& = \frac{1}{{24}}\left[ {\ln \left| {\frac{{4 + 3}}{{4 - 3}}} \right| - \ln \left| {\frac{{4 - 3}}{{4 - 3}}} \right|} \right] \cr
& = \frac{1}{{24}}\left[ {\ln \left| {\frac{7}{1}} \right| - \ln \left| {\frac{1}{7}} \right|} \right] \cr
& = \frac{1}{{24}}\left( {\ln 7 + \ln 7} \right) \cr
& = \frac{1}{{12}}\ln 7 \cr} $$
