First slide

Methods of integration

Question

If $f (x) = \frac{x + 2}{2 x + 3} .$ then $\int {(\frac{f (x)}{x^{2}})}^{1 / 2} d x$
is equal to $\frac{1}{\sqrt{2}} g (\frac{1 + \sqrt{2 f (x)}}{1 - \sqrt{2 f (x)}}) - \sqrt{\frac{2}{3}} h (\frac{\sqrt{3 f (x)} + \sqrt{2}}{\sqrt{3 f (x)} - \sqrt{2}})$
$+ C$ where

Moderate

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

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

Solution

Putting $y^{2} = f (x) = \frac{x + 2}{2 x + 3}$ , we have
$x = \frac{3 y^{2} - 2}{1 - 2 y^{2}}$ and $d x = - \frac{2 y}{{(1 - 2 y^{2})}^{2}} d y .$
So $\begin{array}{l} = - \int y \cdot \frac{2 y}{{(1 - 2 y^{2})}^{2}} \cdot \frac{1 - 2 y^{2}}{3 y^{2} - 2} d y \\ = 2 \int \frac{y^{2}}{(2 y^{2} - 1) (3 y^{2} - 2)} d y \\ = - 2 \int [\frac{1}{2 y^{2} - 1} - \frac{2}{3 y^{2} - 2}] d y \\ = \frac{1}{\sqrt{2}} \log |\frac{1 + \sqrt{2} y}{1 - \sqrt{2} y}| - \sqrt{\frac{2}{3}} \log |\frac{\sqrt{3} y + \sqrt{2}}{\sqrt{3} y - \sqrt{2}}| + C \end{array}$
Thus $g (x) = \log | x | and h (x) = \log | x |$ .

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