Question

please help me with question of integrals.​

Answer 1

Answer:

(d) is a right ans

please make brainlyest ans

Answer 2

$\large\underline{\sf{Given \:Question - }}$

$\sf \: \displaystyle\int_0^3\rm \frac{3x + 1}{ {x}^{2} + 9} \: dx \: =$

$\: \: \: \: \rm \: \: (a). \: \: \: log\bigg[2\sqrt{2} \bigg] \: + \dfrac{\pi}{12}$

$\: \: \: \: \rm \: \: (b). \: \: \: log\bigg[2\sqrt{2} \bigg] \: + \dfrac{\pi}{3}$

$\: \: \: \: \rm \: \: (c). \: \: \: log\bigg[\sqrt{2} \bigg] \: + \dfrac{\pi}{12}$

$\: \: \: \: \rm \: \: (d). \: \: \: log\bigg[2\sqrt{2} \bigg] \: + \dfrac{\pi}{6}$

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

Given integral is

$\rm :\longmapsto\: \: \displaystyle\int_0^3\rm \frac{3x + 1}{ {x}^{2} + 9} \: dx \:$

can be rewritten as

$\rm \: = \: \: \displaystyle\int_0^3\rm \frac{3x}{ {x}^{2} + 9} \: dx \: + \displaystyle\int_0^3\rm \frac{1}{ {x}^{2} + 9} \: dx$

can be further rewritten as

$\rm \: = \:\dfrac{3}{2} \: \displaystyle\int_0^3\rm \frac{2x}{ {x}^{2} + 9} \: dx \: + \displaystyle\int_0^3\rm \frac{1}{ {x}^{2} + {3}^{2} } \: dx$

We know,

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

and

$\boxed{ \bf{ \: \displaystyle\int\rm \frac{f'(x)}{f(x)} dx = logf(x) + c}}$

So, using these results, we get

$\rm \: = \:\dfrac{3}{2} \bigg[log( {x}^{2} + 9)\bigg]_0^3 \: + \dfrac{1}{3}\bigg[ {tan}^{ - 1} \dfrac{x}{3}\bigg]_0^3$

$\rm \: = \:\dfrac{3}{2} \bigg[log( 9 + 9) - log9\bigg] \: + \dfrac{1}{3}\bigg[ {tan}^{ - 1} \dfrac{3}{3} - {tan}^{ - 1} 0\bigg]$

$\rm \: = \:\dfrac{3}{2} \bigg[log(18) - log9\bigg] \: + \dfrac{1}{3}\bigg[ {tan}^{ - 1} 1 - 0\bigg]$

We know,

$\boxed{ \bf{ \: log(x) - log(y) = log\bigg( \frac{x}{y}\bigg) }}$

So, using this,

$\rm \: = \:\dfrac{3}{2} \bigg[log \dfrac{18}{9} \bigg] \: + \dfrac{1}{3}\bigg[ \dfrac{\pi}{4} \bigg]$

$\rm \: = \:\dfrac{3}{2} \bigg[log 2 \bigg] \: + \dfrac{\pi}{12}$

We know,

$\boxed{ \bf{ \: y log(x) = log( {x}^{y} )}}$

So, using this result

$\rm \: = \: log\bigg[ 2 \bigg]^{\dfrac{3}{2}} \: + \dfrac{\pi}{12}$

$\rm \: = \: log\bigg[ \sqrt{2} \bigg]^{3} \: + \dfrac{\pi}{12}$

$\rm \: = \: log\bigg[2\sqrt{2} \bigg] \: + \dfrac{\pi}{12}$

Therefore,

$\red{\rm :\longmapsto\: \: \displaystyle\int_0^3\rm \frac{3x + 1}{ {x}^{2} + 9} \: dx \: = \: \: log\bigg[2\sqrt{2} \bigg] \: + \dfrac{\pi}{12} }$

• Hence, Option (a) is correct.

Additional Information :-

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