Answer
$$\frac{{{e^x}}}{{\sqrt {1 - {e^{2x}}} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{{{\left( {1 - {e^{2x}}} \right)}^{3/2}}}}} dx \cr
& {\text{Let }}u = {e^x}.{\text{ Then }}du = {e^x}dx,{\text{ }}dx = \frac{1}{{{e^x}}}du \cr
& {\text{Substituting you have}} \cr
& \int {\frac{{{e^x}}}{{{{\left( {1 - {e^{2x}}} \right)}^{3/2}}}}} dx = \int {\frac{{{e^x}}}{{{{\left( {1 - {u^2}} \right)}^{3/2}}}}\frac{1}{{{e^x}}}} du \cr
& = \int {\frac{1}{{{{\left( {1 - {u^2}} \right)}^{3/2}}}}} du \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{1}{{{{\left( {{a^2} - {u^2}} \right)}^{3/2}}}}} du = \frac{u}{{{a^2}\sqrt {{a^2} - {u^2}} }} + C \cr
& \int {\frac{1}{{{{\left( {1 - {u^2}} \right)}^{3/2}}}}} du = \frac{u}{{\sqrt {1 - {u^2}} }} + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = {e^x} \cr
& = \frac{{{e^x}}}{{\sqrt {1 - {e^{2x}}} }} + C \cr} $$
