Question

Integrate:-


$I= \displaystyle \int \dfrac{(x + 1)}{(x + 2)(x + 3)}dx$

Solve the question and please explain each step properly.​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{x+1}{(x+2)(x+3)}\ dx\quad\quad\dots(1)$

Let,

$\longrightarrow\dfrac{x+1}{(x+2)(x+3)}=\dfrac{A}{x+2}+\dfrac{B}{x+3}$

$\longrightarrow\dfrac{x+1}{(x+2)(x+3)}=\dfrac{A(x+3)+B(x+2)}{(x+2)(x+3)}$

$\longrightarrow x+1=(A+B)x+(3A+2B)$

Equating corresponding coefficients,

• $A+B=1$

• $3A+2B=1$

Solving them, we get,

• $A=-1$

• $B=2$

Then (1) becomes,

$\displaystyle\longrightarrow I=\int\left(\dfrac{-1}{x+2}+\dfrac{2}{x+3}\right)\ dx$

$\displaystyle\longrightarrow I=-\log|x+2|+2\log|x+3|+C$

$\displaystyle\longrightarrow\underline{\underline{I=\log\left|\dfrac{(x+3)^2}{x+2}\right|+C}}$

Answer 2

\

$\int \frac{(x + 1)}{(x + 2)(x + 3)} \\ \frac{(x + 1)}{(x + 2)(x + 3)} = \frac{a}{(x + 2)} + \frac{b}{(x + 3)} \\ \frac{x + 1}{(x +2 )(x + 3)} = \frac{3a + ax + bx + 2b}{(x + 2)(x + 3)} \\ cancell \: out \: denominator \\ x + 1 = x(a + b) + 3a + 2b \\ on \: comparing \: both \: term \: \: we \: get \\ a + b = 1 \: \: (coef.of \: x \: is \: 1) \\ or \: a = 1 - b \\ 3a + 2b = 1 \\ putting \: value \: of \: b \: we \: get \\ 3(1 - b) + 2b = 1 \\ b = 2 \: and \: a = - 1 \\ putting \: the \: value \: we \: get \\ \frac{(x + 1)}{(x + 2)(x + 3)} = \frac{ - 1}{(x + 2)} + \frac{2}{(x + 3)} \\ apply \: integration \: both \: side \\ \int \: \frac{(x + 1)}{(x + 2)(x + 3)} = \int \frac{ - 1}{(x + 2)} + \int \frac{2}{x + 3} \\ \int \: \frac{(x + 1)}{(x + 2)(x + 3)} = - log |x + 2| + 2log |x + 3| + c$