The value of
$\int \frac{x^{5} + x^{4} + 4 x^{3} + 4 x^{2} + 4 x + 4}{{(x^{2} + 2)}^{5}}$ is equal to

Moderate

A

$\frac{4 x - 3}{{(x^{2} + 1)}^{2}} + \frac{3}{8} \frac{x}{x^{2} + 2} + \frac{1}{\sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}} + C$
B

$\frac{1}{12} \frac{2 x - 3}{{(x^{2} + 2)}^{2}} + \frac{3}{16 x^{2} + 2} + \frac{3}{16 \sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}} + C$
C

$\frac{2 x - 3}{{(x^{2} + 2)}^{2}} + \frac{3}{8 x^{2} + 2} + \frac{1}{2 \sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}} + C$
D

none of these

Solution

$\begin{array}{l} x^{5} + x^{4} + 4 x^{3} + 4 x^{2} + 4 x + 4 \\ = x^{3} (x^{2} + 2) + x^{2} (x^{2} + 2) + 2 x (x^{2} + 2) \\ + 2 (x^{2} + 2) \\ = (x^{2} + 2) (x^{3} + x^{2} + 2 x + 2) \\ = (x^{2} + 2) (x (x^{2} + 2) + (x^{2} + 2)) \end{array}$
$= {(x^{2} + 2)}^{2} (x + 2)$ . Therefore,
$\begin{array}{l} f (x) = \frac{x^{5} + x^{4} + 4 x^{3} + 4 x^{2} + 4 x + 4}{{(x^{2} + 2)}^{5}} \\ = \frac{x + 2}{{(x^{2} + 2)}^{3}} = \frac{1}{2} \frac{2 x}{{(x^{2} + 2)}^{3}} + \frac{2}{{(x^{2} + 2)}^{3}} \end{array}$
Now, $\int \frac{d x}{{(x^{2} + 2)}^{3}} = \frac{1}{8} \frac{x}{{(x^{2} + 2)}^{2}} + \frac{3}{8} \int \frac{d x}{{(x^{2} + 2)}^{2}}$
$(using \int \frac{d y}{{(y^{2} + k^{2})}^{n}})$
$\begin{array}{l} \int \frac{d x}{{(x^{2} + 2)}^{2}} = \frac{1}{4} \frac{x}{x^{2} + 2} + \frac{1}{4 \sqrt{2}} = \tan^{- 1} \frac{x}{\sqrt{2}} + C \\ ∴ \int f (x) d x = - \frac{1}{4 {(x^{2} + 2)}^{2}} + \frac{1}{4} {(x^{2} + 2)}^{2} \\ + \frac{3}{4} [\frac{1}{4} \frac{x}{x^{2} + 2} + \frac{1}{4 \sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}}] + C \end{array}$
$= \frac{x - 1}{4 {(x^{2} + 2)}^{2}} + \frac{3}{16} \frac{x}{x^{2} + 2} + \frac{3}{16 \sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}} + C$ .

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