Solution

To simplify the integral, the integrand $\dfrac{3x+2}{(x+1)(x+2)}$ must be split into partial fractions like so:

\[\dfrac{3x+2}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}.\]

This can be simplified by multiplying the equation by $(x+1)(x+2)$:

\begin{align} \dfrac{(3x+2)(x+1)(x+2)}{(x+1)(x+2)} &= \dfrac{A(x+1)(x+2)}{x+1}+\dfrac{B(x+1)(x+2)}{x+2} \\ 3x+2 &= A(x+2)+B(x+1). \end{align}

$A$ is found by setting $x=-1$:

\begin{align} 3\cdot(-1)+2 &= A\bigl((-1)+2\bigl)+B\bigl(x+1\bigl) \\ -3+2 &= A\cdot1+B\cdot0 \\ \Rightarrow A &= -1. \end{align}

Similarly, $B$ is found by setting $x=-2$:

\begin{align} 3\cdot(-2) +2 &= A\bigl((-2)+2\bigl)+B\bigl((-2)+1\bigl) \\ -6+2 &= A\cdot0+B\cdot(-1) \\ \Rightarrow B &= 4. \end{align}

Hence,

\[\dfrac{3x+2}{(x+1)(x+2)}=\dfrac{4}{x+2}-\dfrac{1}{x+1},\]

and the integral becomes:

\[\int\dfrac{3x+2}{(x+1)(x+2)}\;\mathrm{d}x=\int\dfrac{4}{x+2}\;\mathrm{d}x-\int\dfrac{1}{x+1}\;\mathrm{d}x.\]

Recall that the derivative of $\ln \left| f(x) \right|$ is $\dfrac{f'(x)}{f(x)}$. Since differentiation is the opposite of integration (by the fundamental theorem of calculus), this implies:

\[\int\dfrac{f'(x)}{f(x)}\;\mathrm{d}x=\ln{\vert f(x)\vert }+\text{ constant}.\]

Now apply this result to a general function $\dfrac{1}{x+k}$, where $k$ is a constant. Note that the derivative of $x+k$ is $1$. Hence,

\[\int\dfrac{1}{x+k}\;\mathrm{d}x=\ln{\vert x+k\vert }+\text{ constant}.\]

By this result,

\begin{align} \int\dfrac{3x+2}{(x+1)(x+2)}\;\mathrm{d}x &= \int\dfrac{4}{x+2}\;\mathrm{d}x-\int\dfrac{1}{x+1}\;\mathrm{d}x \\ &= 4\int\dfrac{1}{x+2}\;\mathrm{d}x-\int\dfrac{1}{x+1}\;\mathrm{d}x \\ &=4\ln{\vert x+2\vert }-\ln{\vert x+1\vert }+\mathrm{C}, \end{align}

Note that the arbitrary constants of integration arising from each partial fraction have been absorbed into one constant $\mathrm{C}$.

Solution

Note that there is a repeated factor of $x-1$ in the denominator. The partial fraction decomposition is therefore given by:

\[\dfrac{1}{(x-1)^2(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}.\]

This can be simplified by multiplying both sides of the equation by $(x-1)^2(x+1)$:

\begin{align} \dfrac{(x-1)^2(x+1)}{(x-1)^2(x+1)} &= \dfrac{A(x-1)^2(x+1)}{x-1} \\ &\quad +\dfrac{B(x-1)^2(x+1)}{(x-1)^2} \\ &\quad+\dfrac{C(x-1)^2(x+1)}{x+1}\\ 1 &= A(x-1)(x+1)+B(x+1)+C(x-1)^2. \end{align}

$B$ is found by setting $x=1$:

\begin{align} 1 &= A(1-1)(1+1)+B(1+1)+C(1-1)^2 \\ &=A\cdot0+B\cdot2+C\cdot 0 \\ \\ \Rightarrow B &= \dfrac{1}{2} \end{align}

$C$ is found by setting $x=-1$:

\begin{align} 1 &= A(-1-1)(-1+1)+B(-1+1)+C(-1-1)^2 \\ &=A\cdot0+B\cdot0+4C \\ \Rightarrow C &= \dfrac{1}{4} \end{align}

Substituting these values for $B$ and $C$ into the expression and setting $x=0$ gives:

\begin{align} 1 &= A(0-1)(0+1)+\dfrac{1}{2}(0+1)+\dfrac{1}{4}(0-1)^2 \\ &= -A+\dfrac{1}{2}+\dfrac{1}{4} \\ \Rightarrow A &= -\dfrac{1}{4} \end{align}

Hence,

\[\dfrac{1}{(x-1)^2(x+1)}=-\dfrac{1}{4(x-1)}+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\]

and the integral becomes

\[\int\dfrac{1}{(x-1)^2(x+1)}\mathrm{d} x =\int-\dfrac{1}{4(x-1)}+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\mathrm{d} x.\]

This can be evaluated directly, using standard integrals:

\begin{align} \int\dfrac{1}{4(x-1)}&+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\mathrm{d} x \\ &= -\dfrac{1}{4}\ln{\vert x-1\vert }-\dfrac{1}{2(x-1)}+\dfrac{1}{4}\ln{\vert x+1\vert }+\mathrm{C}. \end{align}