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.