## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{9}{\tanh ^{ - 1}}\frac{1}{3}$$
\eqalign{ & \int_0^1 {\frac{{{x^2}}}{{9 - {x^6}}}} dx \cr & {\text{rewriting the denominator}} \cr & = \int_0^1 {\frac{{{x^2}}}{{{{\left( 3 \right)}^2} - {{\left( {{x^3}} \right)}^2}}}} dx \cr & {\text{substitute }}u = {x^3},{\text{ }}du = 3{x^2}dx \cr & {\text{express the limits in terms of }}u \cr & x = 1{\text{ implies }}u = {\left( 1 \right)^3} = 1 \cr & x = 0{\text{ implies }}u = {\left( 0 \right)^3} = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^1 {\frac{{{x^2}}}{{{{\left( 3 \right)}^2} - {{\left( {{x^3}} \right)}^2}}}} dx = \frac{1}{3}\int_0^1 {\frac{{du}}{{{3^2} - {u^2}}}} \cr & {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, }}\left| {{e^x}} \right|{\text{ > 0}} \cr & \int {\frac{{dx}}{{{a^2} - {u^2}}}} = \frac{1}{a}{\tanh ^{ - 1}}\frac{u}{a} + C \cr & = \left. {\frac{1}{3}\left( {\frac{1}{3}{{\tanh }^{ - 1}}\frac{u}{3}} \right)} \right|_0^1 \cr & = \frac{1}{9}\left( {{{\tanh }^{ - 1}}\frac{1}{3}} \right) - \frac{1}{9}\left( {{{\tanh }^{ - 1}}0} \right) \cr & {\text{simplify}} \cr & = \frac{1}{9}{\tanh ^{ - 1}}\frac{1}{3} \cr}