Answer
$$ - \frac{1}{{22}}\ln \left| {\cos 11{x^2}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {x\tan 11{x^2}} dx \cr
& {\text{set }}u = 11{x^2}{\text{ then }}\frac{{du}}{{dx}} = 22x,\,\,\,\,\,\,\,\,\frac{{du}}{{22x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {x\tan 11{x^2}} dx = \int {x\tan u} \left( {\frac{{du}}{{22x}}} \right) \cr
& {\text{cancel the common factor }}{e^{ - x}} \cr
& = \int {\tan u} \left( {\frac{{du}}{{22}}} \right) \cr
& = \frac{1}{{22}}\int {\tan u} du \cr
& {\text{ using the Basic Trigonometric integral }}\int {\tan x} dx = - \ln \left| {\cos x} \right| + C{\text{ }}\left( {{\text{see page 694}}} \right) \cr
& = \frac{1}{{22}}\left( { - \ln \left| {\cos u} \right|} \right) + C \cr
& = - \frac{1}{{22}}\ln \left| {\cos u} \right| + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{{22}}\ln \left| {\cos 11{x^2}} \right| + C \cr} $$
