By using the substitution \(u = {{\text{e}}^x} + 3\), find \(\int {\frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 6{{\text{e}}^x} + 13}}{\text{d}}x} \).

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^x}\)     (A1)

EITHER

integral is \(\int {\frac{{{{\text{e}}^x}}}{{{{({{\text{e}}^x} + 3)}^2} + {2^2}}}{\text{d}}x} \)     M1A1

\( = \frac{1}{{{u^2} + {2^2}}}{\text{d}}u\)     M1A1

Note:     Award M1 only if the integral has completely changed to one in \(u\).

Note:     \({\text{d}}u\) needed for final A1

OR

\({{\text{e}}^x} = u - 3\)

integral is \(\int {\frac{1}{{{{(u - 3)}^2} + 6(u - 3) + 13}}{\text{d}}u} \)     M1A1

Note: Award M1 only if the integral has completely changed to one in \(u\).

\( = \int {\frac{1}{{{u^2} + {2^2}}}{\text{d}}u} \)     M1A1

Note:     In both solutions the two method marks are independent.

THEN

\( = \frac{1}{2}\arctan \left( {\frac{u}{2}} \right)( + c)\)     (A1)

\( = \frac{1}{2}\arctan \left( {\frac{{{{\text{e}}^x} + 3}}{2}} \right)( + c)\)     A1

Total [7 marks]

Many good complete answers. Some did not realise it was arctan. Some had poor understanding of the method.