Question

evaluate:-
int dx/ (x+2)(x^ 2 +1) = integrate 1/(x ^ 2 + 1) dx = arctan(x) + c​

Answer 1

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

Given integral is

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

Using Partial Fraction, Let assume that

$\rm \: \dfrac{1}{(x + 2)( {x}^{2} + 1)} = \dfrac{A}{x + 2} + \dfrac{Bx + C}{ {x}^{2} + 1} \cdots(1)$

On taking LCM, we get

$\rm \: 1 = A( {x}^{2} + 1) + (Bx + C)(x + 2) \cdots \cdots(2)$

On substituting x = - 2, we get

$\rm \: 1 = A( {( - 2)}^{2} + 1) + 0$

$\rm \: 1 = A( 4+ 1)$

$\rm\implies \:\boxed{ \rm{ \:A = \frac{1}{5} \: }}$

Now, from equation (2), we have

$\rm \: 1 = A{x}^{2} + A + {Bx}^{2} + Cx + 2Bx + 2C$

$\rm \: 1 = (A + B){x}^{2} + (C + 2B)x + (A + 2C)$

On comparing the coefficient of x^2, we have

$\rm \: A + B = 0$

$\rm \: B = - A$

$\rm\implies \:\boxed{ \rm{ \:B = - \frac{1}{5} \: }}$

Now, On comparing the constant term, we get

$\rm \: A + 2C = 1$

$\rm \: \frac{1}{5} + 2C = 1$

$\rm \: 2C = 1 - \frac{1}{5}$

$\rm \: 2C = \frac{4}{5}$

$\rm\implies \:\boxed{ \rm{ \:C = \frac{2}{5} \: }}$

So, equation (1) can be rewritten as

$\rm \: \dfrac{1}{(x + 2)( {x}^{2} + 1)} = \dfrac{A}{x + 2} + \dfrac{Bx}{ {x}^{2} + 1} + \frac{C}{ {x}^{2} + 1}$

On substituting the values of A, B and C, we get

$\rm \: \dfrac{1}{(x + 2)( {x}^{2} + 1)} = \dfrac{ \dfrac{1}{5} }{x + 2} + \dfrac{ - \dfrac{1}{5} x}{ {x}^{2} + 1} + \frac{ \dfrac{2}{5} }{ {x}^{2} + 1}$

On integrating both sides, we get

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

$\rm \:\displaystyle\int\rm \dfrac{dx}{(x + 2)( {x}^{2} + 1)} = \frac{1}{5}log |x + 2|-\frac{1}{10}\displaystyle\int\rm \dfrac{2xdx}{ {x}^{2} + 1} +\frac{2}{5} {tan}^{ - 1}x$

$\displaystyle\int\rm\dfrac{dx}{(x + 2)({x}^{2} + 1)}= \frac{1}{5}log |x + 2|-\frac{1}{10}log | {x}^{2}+ 1|+\frac{2}{5} {tan}^{ - 1}x+c$

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

Formulae Used :-

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

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

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

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Fraction & \bf Partial \:  Fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)}  & \sf \dfrac{p}{ax + b}  + \dfrac{q}{cx + d}  \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} }  & \sf \dfrac{p}{ax + b}  + \dfrac{q}{ {(ax + b)}^{2} }  \\ \\ \sf \dfrac{1}{(ax + b)(c {x}^{2}  + d)}  & \sf \dfrac{p}{ax + b}  + \dfrac{qx + r}{c {x}^{2}  + d}  \end{array}} \\ \end{gathered}$

Answer 2

Answer:

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

Solution−

Given integral is

\begin{gathered}\rm \: \displaystyle\int\rm \frac{1}{(x + 2)( {x}^{2} + 1)} dx \\ \end{gathered}

∫

(x+2)(x

2

+1)

1

dx

Using Partial Fraction, Let assume that

\begin{gathered}\rm \: \dfrac{1}{(x + 2)( {x}^{2} + 1)} = \dfrac{A}{x + 2} + \dfrac{Bx + C}{ {x}^{2} + 1} \cdots(1) \\ \end{gathered}

(x+2)(x

2

+1)

1

=

x+2

A

+

x

2

+1

Bx+C

⋯(1)

On taking LCM, we get

\begin{gathered}\rm \: 1 = A( {x}^{2} + 1) + (Bx + C)(x + 2) \cdots \cdots(2)\\ \end{gathered}

1=A(x

2

+1)+(Bx+C)(x+2)⋯⋯(2)

On substituting x = - 2, we get

\begin{gathered}\rm \: 1 = A( {( - 2)}^{2} + 1) + 0\\ \end{gathered}

1=A((−2)

2

+1)+0

\begin{gathered}\rm \: 1 = A( 4+ 1)\\ \end{gathered}

1=A(4+1)

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:A = \frac{1}{5} \: }} \\ \end{gathered}

⟹

A=

5

1

Now, from equation (2), we have

\begin{gathered}\rm \: 1 = A{x}^{2} + A + {Bx}^{2} + Cx + 2Bx + 2C \\ \end{gathered}

1=Ax

2

+A+Bx

2

+Cx+2Bx+2C

\begin{gathered}\rm \: 1 = (A + B){x}^{2} + (C + 2B)x + (A + 2C) \\ \end{gathered}

1=(A+B)x

2

+(C+2B)x+(A+2C)

On comparing the coefficient of x^2, we have

\begin{gathered}\rm \: A + B = 0 \\ \end{gathered}

A+B=0

\begin{gathered}\rm \: B = - A \\ \end{gathered}

B=−A

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:B = - \frac{1}{5} \: }} \\ \end{gathered}

⟹

B=−

5

1

Now, On comparing the constant term, we get

\begin{gathered}\rm \: A + 2C = 1 \\ \end{gathered}

A+2C=1

\begin{gathered}\rm \: \frac{1}{5} + 2C = 1 \\ \end{gathered}

5

1

+2C=1

\begin{gathered}\rm \: 2C = 1 - \frac{1}{5} \\ \end{gathered}

2C=1−

5

1

\begin{gathered}\rm \: 2C = \frac{4}{5} \\ \end{gathered}

2C=

5

4

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:C = \frac{2}{5} \: }} \\ \end{gathered}

⟹

C=

5

2

So, equation (1) can be rewritten as

\begin{gathered}\rm \: \dfrac{1}{(x + 2)( {x}^{2} + 1)} = \dfrac{A}{x + 2} + \dfrac{Bx}{ {x}^{2} + 1} + \frac{C}{ {x}^{2} + 1} \\ \end{gathered}

(x+2)(x

2

+1)

1

=

x+2

A

+

x

2

+1

Bx

+

x

2

+1

C

On substituting the values of A, B and C, we get

\begin{gathered}\rm \: \dfrac{1}{(x + 2)( {x}^{2} + 1)} = \dfrac{ \dfrac{1}{5} }{x + 2} + \dfrac{ - \dfrac{1}{5} x}{ {x}^{2} + 1} + \frac{ \dfrac{2}{5} }{ {x}^{2} + 1} \\ \end{gathered}

(x+2)(x

2

+1)

1

=

x+2

5

1

+

x

2

+1

−

5

1

x

+

x

2

+1

5

2

On integrating both sides, we get

\begin{gathered}\rm \:\displaystyle\int\rm \dfrac{dx}{(x + 2)( {x}^{2} + 1)} = \frac{1}{5}\displaystyle\int\rm \dfrac{dx}{x + 2} - \frac{1}{5}\displaystyle\int\rm \dfrac{xdx}{ {x}^{2} + 1} + \frac{2}{5}\displaystyle\int\rm \frac{dx}{ {x}^{2} + 1} \\ \end{gathered}

∫

(x+2)(x

2

+1)

dx

=

5

1

∫

x+2

dx

−

5

1

∫

x

2

+1

xdx

+

5

2

∫

x

2

+1

dx

\begin{gathered}\rm \:\displaystyle\int\rm \dfrac{dx}{(x + 2)( {x}^{2} + 1)} = \frac{1}{5}log |x + 2|-\frac{1}{10}\displaystyle\int\rm \dfrac{2xdx}{ {x}^{2} + 1} +\frac{2}{5} {tan}^{ - 1}x \\ \end{gathered}

∫

(x+2)(x

2

+1)

dx

=

5

1

log∣x+2∣−

10

1

∫

x

2

+1

2xdx

+

5

2

tan

−1

x

\begin{gathered}\displaystyle\int\rm\dfrac{dx}{(x + 2)({x}^{2} + 1)}= \frac{1}{5}log |x + 2|-\frac{1}{10}log | {x}^{2}+ 1|+\frac{2}{5} {tan}^{ - 1}x+c\\ \end{gathered}

∫

(x+2)(x

2

+1)

dx

=

5

1

log∣x+2∣−

10

1

log∣x

2

+1∣+

5

2

tan

−1

x+c

\rule{190pt}{2pt}

Formulae Used :-

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

∫

x

dx

=log∣x∣+c

\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

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf Fraction & \bf Partial \: Fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)} & \sf \dfrac{p}{ax + b} + \dfrac{q}{cx + d} \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} } & \sf \dfrac{p}{ax + b} + \dfrac{q}{ {(ax + b)}^{2} } \\ \\ \sf \dfrac{1}{(ax + b)(c {x}^{2} + d)} & \sf \dfrac{p}{ax + b} + \dfrac{qx + r}{c {x}^{2} + d} \end{array}} \\ \end{gathered}\end{gathered}

Fraction

(ax+b)(cx+d)

1

(ax+b)

2

1

(ax+b)(cx

2

+d)

1

Partial Fraction

ax+b

p

+

cx+d

q

ax+b

p

+

(ax+b)

2

q

ax+b

p

+

cx

2

+d

qx+r

$\leqslant answer > \geqslant$