Question

evaluate 1/ (x+1) (x+2) (x+3) dx​

Answer 1

Solution:

Given Integral:

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

To solve this, we have to decompose it into partial fraction (See attachment)

After decomposing to partial fractions, we get:

$\displaystyle \rm \longrightarrow I = \int \bigg[ \dfrac{1}{2(x + 1)} - \dfrac{1}{x + 2} + \dfrac{1}{2(x + 3)} \bigg] \: dx$

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

$\displaystyle \rm \longrightarrow I = \dfrac{1}{2} \int \dfrac{1}{x + 1} \: dx - \int\dfrac{1}{x + 2} \: dx + \dfrac{1}{2} \int \dfrac{1}{x + 3} \: dx$

Let us assume that:

$\rm \longrightarrow u =x + 1$

$\rm \longrightarrow v =x + 2$

$\rm \longrightarrow w =x + 3$

Therefore:

$\rm \longrightarrow du =dx$

$\rm \longrightarrow dv =dx$

$\rm \longrightarrow dw =dx$

So, the integral changes to:

$\displaystyle \rm \longrightarrow I = \dfrac{1}{2} \int \dfrac{du}{u} - \int\dfrac{dv}{v} + \dfrac{1}{2} \int \dfrac{dw}{w}$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{2} \ln( |u| ) - \ln( |v| )+ \dfrac{1}{2} \ln( |w| ) + C$

Substituting back the values of u, v and w, we get:

$\displaystyle \rm \longrightarrow I = \dfrac{1}{2} \ln( |x + 1| ) - \ln( |x + 2| )+ \dfrac{1}{2} \ln( |x +3 | ) + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{dx}{(x + 1)(x + 2)(x + 3)} = \dfrac{1}{2} \ln( |x + 1| ) - \ln( |x + 2| )+ \dfrac{1}{2} \ln( |x +3 | ) + C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$