Question

Integrate with respect to x
$\frac{ {x}^{2} + 4}{ {x}^{2} + 16 }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{2} + 4}{ {x}^{4} + 16} dx$

To evaluate this integral,

$\rm :\longmapsto\:Divide \: numerator \: and \: denominator \: of \: {x}^{2}$

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{4}{ {x}^{2} } }{ {x}^{2} + \dfrac{16}{ {x}^{2} } } \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{4}{ {x}^{2} } }{ {x}^{2} + {\bigg[\dfrac{4}{x} \bigg]}^{2} } \: dx$

We know,

$\boxed{ \tt{ \: {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \: }}$

So, using this identity, in denominator, we get

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{4}{ {x}^{2} } }{ {\bigg[x - \dfrac{4}{x} \bigg]}^{2} + 2 \times x \times \dfrac{4}{x} } \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{1 + \dfrac{4}{ {x}^{2} } }{ {\bigg[x - \dfrac{4}{x} \bigg]}^{2} + 8 } \: dx$

Now, we Substitute

$\red{\rm :\longmapsto\:x - \dfrac{4}{x} = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:1 + \dfrac{4}{ {x}^{2} } = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:\bigg(1 + \dfrac{4}{ {x}^{2} }\bigg)dy = dx}$

So, on substituting all these values, we get

$\rm \: = \: \displaystyle\int\rm \frac{dy}{ {y}^{2} + 8}$

$\rm \: = \: \displaystyle\int\rm \frac{dy}{ {y}^{2} + {(2 \sqrt{2}) }^{2} }$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c \: }}$

So, using this, we get

$\rm \: = \: \dfrac{1}{2 \sqrt{2} } {tan}^{ - 1} \dfrac{y}{2 \sqrt{2} } + c$

$\rm \: = \: \dfrac{1}{2 \sqrt{2} } {tan}^{ - 1} \dfrac{x - \dfrac{4}{x} }{2 \sqrt{2} } + c$

$\rm \: = \: \dfrac{1}{2 \sqrt{2} } {tan}^{ - 1} \dfrac{\dfrac{ {x}^{2} - 4}{x} }{2 \sqrt{2} } + c$

$\rm \: = \: \dfrac{1}{2 \sqrt{2} } {tan}^{ - 1} \dfrac{ {x}^{2} - 4 }{2 \sqrt{2} x} + c$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$