By using the substitution $x = 2\tan u$, show that $\int {\frac{{{\text{d}}x}}{{{x^2}\sqrt {{x^2} + 4} }} = \frac{{ - \sqrt {{x^2} + 4} }}{{4x}} + C}$.

EITHER

$\frac{{{\text{d}}x}}{{{\text{d}}u}} = 2\,{\text{se}}{{\text{c}}^2}u$     A1

$\int {\frac{{2\,{\text{se}}{{\text{c}}^2}u{\text{d}}u}}{{4{{\tan }^2}u\sqrt {4 + 4{{\tan }^2}u} }}}$     (M1)

$\int {\frac{{2\,{\text{se}}{{\text{c}}^2}u{\text{d}}u}}{{4{{\tan }^2}u \times 2\,\sec u}}}$   $( = \int {\frac{{{\text{d}}u}}{{4{{\sin }^2}u\sqrt {{{\tan }^2}u + 1} }}{\text{ or }} = \int {\frac{{2\,{\text{se}}{{\text{c}}^2}u{\text{d}}u}}{{4{{\tan }^2}u\sqrt {4{{\sec }^2}u} }})} }$     A1

OR

$u = \arctan \frac{x}{2}$

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{2}{{{x^2} + 4}}$     A1

$\int {\frac{{\sqrt {4{{\tan }^2}u + 4{\text{d}}u} }}{{2 \times 4{{\tan }^2}u}}}$     (M1)

$\int {\frac{{2\,\sec u{\text{d}}u}}{{2 \times 4{{\tan }^2}u}}}$     A1

THEN

$= \frac{1}{4}\int {\frac{{\sec u{\text{d}}u}}{{{{\tan }^2}u}}}$

$= \frac{1}{4}\int {{\text{cosec}}\,u\cot u{\text{d}}u{\text{ }}\left( { = \frac{1}{4}\int {\frac{{\cos u}}{{{{\sin }^2}u}}{\text{d}}u} } \right)}$     A1

$=  - \frac{1}{4}{\text{cosec}}\,u( + C){\text{ }}\left( { =  - \frac{1}{{4\sin u}}( + C)} \right)$     A1

use of either $u = \frac{x}{2}$ or an appropriate trigonometric identity     M1

either $\sin u = \frac{x}{{\sqrt {{x^2} + 4} }}$ or ${\text{cosec}}\,u = \frac{{\sqrt {{x^2} + 4} }}{x}$ (or equivalent)     A1

$= \frac{{ - \sqrt {{x^2} + 4} }}{{4x}}( + C)$     AG

[7 marks]

Most candidates found this a challenging question. A large majority of candidates were able to change variable from x to u but were not able to make any further progress.