Question

Evaluate :-

${\bold{ \sf{ \int \frac{x^{2} }{(x^{2} + 4)(x ^{2} + 9) } dx}}}$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{I=\displaystyle\int\dfrac{x^2}{(x^2+4)(x^2+9)}\,dx}$

$\bold{\sf{\bullet\,\,Trick\,\,\,for\,\,\,solving\,\,\,Partial\,\,\,Fractions}}$

Consider,

$\sf{\dfrac{t}{(t+1)(t-2)}=\dfrac{A}{t+1}+\dfrac{B}{t-2}}$

To find the coefficient A

Put t= - 1 in the given expression other than (t+1)

i.e.,

$\sf{\dfrac{(-1)}{(t+1)((-1)-2)}=\dfrac{A}{t+1}}$

$\sf{\implies\,\dfrac{A}{t+1}=\dfrac{-1}{(t+1)(-3)}}$

$\sf{\implies\,\dfrac{A}{t+1}=\dfrac{1}{3(t+1)}}$

So, A=1/3

In a similar way, to find B, put t=2 i.e., (t-2=0)

Using the above trick

$\tt{I=\displaystyle\int\dfrac{-4}{(x^2+4)(-4+9)}\,dx+\int\dfrac{-9}{(x^2+9)(-9+4)}\,dx}$

$\tt{\implies\,I=\displaystyle\int\dfrac{-4}{(x^2+4)(5)}\,dx+\int\dfrac{-9}{(x^2+9)(-5)}\,dx}$

$\tt{\implies\,I=\displaystyle-\dfrac{4}{5}\int\dfrac{1}{x^2+4}\,dx+\dfrac{9}{5}\int\dfrac{1}{x^2+9}\,dx}$

$\tt{\implies\,I=\displaystyle-\dfrac{4}{5}\int\dfrac{1}{(x)^2+(2)^2}\,dx+\dfrac{9}{5}\int\dfrac{1}{(x)^2+(3)^2}\,dx}$

$\tt{\implies\,I=\displaystyle-\dfrac{4}{5}\cdot\dfrac{1}{2}\,tan^{-1}\left(\dfrac{x}{2}\right)+\dfrac{9}{5}\cdot\dfrac{1}{3}tan^{-1}\left(\dfrac{x}{3}\right)+C}$

$\tt{\implies\,I=\displaystyle-\dfrac{2}{5}\,tan^{-1}\left(\dfrac{x}{2}\right)+\dfrac{3}{5}tan^{-1}\left(\dfrac{x}{3}\right)+C}$

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{2} }{( {x}^{2} + 4)( {x}^{2} + 9)} \: dx$

can be rewritten as on multiply and divide by 5,

$\rm \: = \: \dfrac{1}{5} \:\displaystyle\int\rm \frac{ 5{x}^{2} }{( {x}^{2} + 4)( {x}^{2} + 9)} \: dx$

$\rm \: = \: \dfrac{1}{5} \:\displaystyle\int\rm \frac{ 9{x}^{2} - 4 {x}^{2} }{( {x}^{2} + 4)( {x}^{2} + 9)} \: dx$

On adding and Subtracting 36,we get

$\rm \: = \: \dfrac{1}{5} \:\displaystyle\int\rm \frac{ 9{x}^{2} - 4 {x}^{2} + 36 - 36}{( {x}^{2} + 4)( {x}^{2} + 9)} \: dx$

$\rm \: = \: \dfrac{1}{5} \:\displaystyle\int\rm \frac{ (9{x}^{2} + 36) - (4{x}^{2} + 36)}{( {x}^{2} + 4)( {x}^{2} + 9)} \: dx$

$\rm \: = \: \dfrac{1}{5} \:\displaystyle\int\rm \frac{ 9({x}^{2} + 4) - 4({x}^{2} + 9)}{( {x}^{2} + 4)( {x}^{2} + 9)} \: dx$

$\rm \: = \: \dfrac{9}{5}\displaystyle\int\rm \frac{dx}{ {x}^{2} + 9} - \dfrac{4}{5}\displaystyle\int\rm \frac{dx}{ {x}^{2} + 4}$

$\rm \: = \: \dfrac{9}{5}\displaystyle\int\rm \frac{dx}{ {x}^{2} + {3}^{2} } - \dfrac{4}{5}\displaystyle\int\rm \frac{dx}{ {x}^{2} + {2}^{2} }$

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

$\rm \: = \: \dfrac{9}{5} \times \dfrac{1}{3} {tan}^{ - 1}\dfrac{x}{3} - \dfrac{4}{5} \times \dfrac{1}{2} {tan}^{ - 1}\dfrac{x}{2} + c$

$\rm \: = \: \dfrac{3}{5} {tan}^{ - 1}\dfrac{x}{3} - \dfrac{2}{5} {tan}^{ - 1}\dfrac{x}{2} + c$

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More to know :-

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