Question

$\bf \displaystyle \int \bf\frac{1}{16 + {x}^{2} } .dx = \frac{1}{P} { \: tan}^{ - 1} \bigg( \frac{x}{Q} \bigg) + C \\ \\ \bf \: then \: find \: (P + Q)$

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Answer 1

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

Given that,

$\rm \: \displaystyle \int \rm\frac{1}{16 + {x}^{2} }dx = \frac{1}{P} { \: tan}^{ - 1} \bigg( \frac{x}{Q} \bigg) + C$

Now, Consider

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

To evaluate this integral, we use Method of Substitution

So, Substitute

$\rm \: x = 4 \: tany \: \rm\implies \:y \: = \: {tan}^{ - 1}\dfrac{x}{4}$

So,

$\rm \: dx = 4 \: {sec}^{2}y \: dy$

So, on substituting these values, we get

$\rm \: = \: \displaystyle \int \rm \: \frac{4 \: {sec}^{2}y \: dy }{16 + 16 {tan}^{2} y}$

$\rm \: = \: \displaystyle \int \rm \: \frac{4 \: {sec}^{2}y \: dy }{16(1 + {tan}^{2} y)}$

$\rm \: = \:\dfrac{1}{4} \displaystyle \int \rm \: \frac{\: {sec}^{2}y \: dy }{{sec}^{2} y}$

$\rm \: = \:\dfrac{1}{4} \displaystyle \int \rm \: dy$

$\rm \: = \:\dfrac{1}{4}y \: + \: c$

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

Thus,

$\rm \: \displaystyle \int \rm\frac{1}{16 + {x}^{2} } dx = \frac{1}{4} { \: tan}^{ - 1} \bigg( \frac{x}{4} \bigg) + C$

As, it is given that

$\rm \: \displaystyle \int \rm\frac{1}{16 + {x}^{2} }dx = \frac{1}{P} { \: tan}^{ - 1} \bigg( \frac{x}{Q} \bigg) + C$

So, on comparing we get

$\rm\implies \:P = 4 \: \: and \: \: Q = 4$

So,

$\rm\implies \:P + Q = 4 + 4 = 8$

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SHORT CUT TRICK

We know

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

So, using this result, we get

$\rm \: \displaystyle \int \rm\frac{1}{16 + {x}^{2} } dx = \frac{1}{4} { \: tan}^{ - 1} \bigg( \frac{x}{4} \bigg) + C$

As it is given that

$\rm \: \displaystyle \int \rm\frac{1}{16 + {x}^{2} }dx = \frac{1}{P} { \: tan}^{ - 1} \bigg( \frac{x}{Q} \bigg) + C$

So, on comparing, we get

$\rm\implies \:P = 4 \: \: and \: \: Q = 4$

So,

$\rm\implies \:P + Q = 4 + 4 = 8$

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ADDITIONAL INFORMATION

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