Question

Evaluate integral (x²+2) / (x3+ 6x + 9) dx​

Answer 1

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

$\rm \: \displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9} \: dx$

To evaluate this integral, we use method of Substitution.

So, substitute

$\rm \: {x}^{3} + 6x + 9 = y$

$\rm \: {3x}^{2} + 6 = \dfrac{dy}{dx}$

$\rm \: 3({x}^{2} + 2) = \dfrac{dy}{dx}$

$\rm \: ({x}^{2} + 2)dx = \dfrac{dy}{3}$

So, on substituting, the above integral, can be rewritten as

$\rm \: = \: \dfrac{1}{3}\displaystyle\int\rm \frac{dy}{y}$

$\rm \: = \: \dfrac{1}{3} \: log |y| + c$

$\rm \: = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c$

Hence,

$\boxed{ \rm{\displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9}dx = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c}}$

$\rule{190pt}{2pt}$

Short Cut Trick :-

$\rm \: \displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9} \: dx$

can be rewritten as

$\rm \: = \dfrac{1}{3} \displaystyle\int\rm \frac{3 {x}^{2} + 6}{ {x}^{3} + 6x + 9} \: dx$

We know,

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

So, using this, we get

$\rm \: = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c$

Hence,

$\boxed{ \rm{\displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9}dx = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c}}$

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

Step-by-step explanation:

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

Solution−

\begin{gathered}\rm \: \displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9} \: dx \\ \end{gathered}

∫

x

3

+6x+9

x

2

+2

dx

To evaluate this integral, we use method of Substitution.

So, substitute

\begin{gathered}\rm \: {x}^{3} + 6x + 9 = y \\ \end{gathered}

x

3

+6x+9=y

\begin{gathered}\rm \: {3x}^{2} + 6 = \dfrac{dy}{dx} \\ \end{gathered}

3x

2

+6=

dx

dy

\begin{gathered}\rm \: 3({x}^{2} + 2) = \dfrac{dy}{dx} \\ \end{gathered}

3(x

2

+2)=

dx

dy

\begin{gathered}\rm \: ({x}^{2} + 2)dx = \dfrac{dy}{3} \\ \end{gathered}

(x

2

+2)dx=

3

dy

So, on substituting, the above integral, can be rewritten as

\rm \: = \: \dfrac{1}{3}\displaystyle\int\rm \frac{dy}{y} =

3

1

∫

y

dy

\begin{gathered}\rm \: = \: \dfrac{1}{3} \: log |y| + c \\ \end{gathered}

=

3

1

log∣y∣+c

\begin{gathered}\rm \: = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c \\ \end{gathered}

=

3

1

log∣x

3

+6x+9∣+c

Hence,

\begin{gathered}\boxed{ \rm{\displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9}dx = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c}} \\ \end{gathered}

∫

x

3

+6x+9

x

2

+2

dx=

3

1

log∣x

3

+6x+9∣+c

\rule{190pt}{2pt}

Short Cut Trick :-

\rm \: \displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9} \: dx∫

x

3

+6x+9

x

2

+2

dx

can be rewritten as

\rm \: = \dfrac{1}{3} \displaystyle\int\rm \frac{3 {x}^{2} + 6}{ {x}^{3} + 6x + 9} \: dx=

3

1

∫

x

3

+6x+9

3x

2

+6

dx

We know,

\begin{gathered}\boxed{ \rm{ \:\displaystyle\int\rm \frac{f'(x)}{f(x)}dx \: = \: log |f(x)| + c \: }} \\ \end{gathered}

∫

f(x)

f

′

(x)

dx=log∣f(x)∣+c

So, using this, we get

\begin{gathered}\rm \: = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c \\ \end{gathered}

=

3

1

log∣x

3

+6x+9∣+c

Hence,

\begin{gathered}\boxed{ \rm{\displaystyle\int\rm \frac{ {x}^{2} + 2}{ {x}^{3} + 6x + 9}dx = \: \dfrac{1}{3} \: log | {x}^{3} + 6x + 9| + c}} \\ \end{gathered}

∫

x

3

+6x+9

x

2

+2

dx=

3

1

log∣x

3

+6x+9∣+c

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\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}\end{gathered}

f(x)

k

sinx

cosx

sec

2

x

cosec

2

x

secxtanx

cosecxcotx

tanx

x

1

e

x

∫f(x)dx

kx+c

−cosx+c

sinx+c

tanx+c

−cotx+c

secx+c

−cosecx+c

logsecx+c

logx+c

e

x

+c