Question

If ƒ(x) = $\frac{x + 2}{2x + 3}$.Then $\int_{}^{}\left( \frac{ƒ(x)}{x^{2}} \right)^{1/2}$dx is equal to

$\frac { 1 } { \sqrt { 2 } }$ $\left( \frac{1 + \sqrt{2ƒ(x)}}{1 - \sqrt{2ƒ(x)}} \right)$–$\sqrt{\frac{2}{3}}$h$\left( \frac{\sqrt{3ƒ(x)} + \sqrt{2}}{\sqrt{3ƒ(x) - \sqrt{2}}} \right)$+ c where –

• A. g(x) = tan^–1 x, h(x) = log |x|
• B. g(x) = log |x|, h(x) = tan^–1 x
• C. g(x) = h(x) = tan^–1 x
• D. g(x) = log |x|, h (x) = log |x|

Answer 1

Answer: (D)

g(x) = log |x|, h (x) = log |x|

Answer 2

Putting y² = ƒ(x) = $\frac{x + 2}{2x + 3}$, we have

x = $\frac { 3 y ^ { 2 } - 2 } { 1 - 2 y ^ { 2 } }$ and dx = – $\frac{2y}{(1–2y^{2})^{2}}$dy

\$\frac{dx}{x}$= –$\int_{}^{}y\frac{2y}{(1 - 2y^{2})^{2}}$. $\frac{1 - 2y^{2}}{3y^{2} - 2}$dy

= 2 $\int_{}^{}\frac{y^{2}}{(2y^{2} - 1)(3y^{2} - 2)}$dy

= –2 $\int_{}^{}\left\lbrack \frac{1}{2y^{2} - 1} - \frac{2}{3y^{2} - 2} \right\rbrack$dy

= $\frac { 1 } { \sqrt { 2 } }$ log $\left| \frac{1 + \sqrt{2}y}{1 - \sqrt{2}y} \right|$– $\sqrt{\frac{2}{3}}$log $\left| \frac{\sqrt{3}y + \sqrt{2}}{\sqrt{3}y - \sqrt{2}} \right|$+ c

Thus g(x) = log |x| and h (x) = log |x|