Example 23.12.

1.

\(\,\displaystyle {\int \frac {x\,dx}{x^2 + 3x + 2}}\)

\[\implies \quad \int \frac {x\,dx}{x^2 + 3x + 2} = \int \frac {x\,dx}{\left ( x + 1\right )\left ( x + 2\right )}\]

\[\frac {x}{\left (x + 2\right )\left (x + 1\right )} = \frac {A}{x + 2} + \frac {B}{x + 1}\]

\begin {align*} \implies \quad x & = A(x+1) + B(x+ 2)\\ x & = Ax + A + Bx + 2B \end {align*}

\begin {align*} A + 2B & = 0\quad \cdots \quad (1)\\ A + B & = 1\quad \cdots \quad (2) \end {align*}

\begin {align*} (1)\quad \implies \quad A & = -2B\\ -2B + B & = 1\quad \implies \quad -B = 1\implies B = -1\\\\ \text {thus}\quad A = -2(-1) & = 2 \end {align*}

\begin {align*} \text {Hence},\qquad \int \frac {x\,dx}{x^2 + 3x + 2} & = \int \frac {2}{x + 2} \,dx - \int \frac {dx}{x + 1}\\\\ & = 2\ln \left ( x + 2\right ) - \ln \left (x + 1\right ) + c\\ & = \ln \left (x + 2\right )^2 - \ln \left (x + 1\right ) + c\\\\ & = \ln \left [\frac {\left (x + 2\right )^2}{x + 1}\right ] + c\\\\ \end {align*}

2.

\(\, \displaystyle {\int \frac {x^3 + 1}{\left (x^2 + 1\right )\left (x - 2\right )}\,dx = \int \frac {x^3 + 1}{x^3 - x^2 - 2}\,dx}\)

So \[\begin {array}{r@{\;}l} & 1\\ x^{3}-x^{2}-2\,\big ) & \overline {x^{3}+0x^{2}+0x+1}\\ & \underline {x^{3}-x^{2}\phantom {{}+0x}-2}\\ & \phantom {x^{3}}x^{2}+0x+3 \end {array}\] that is, the quotient is \(1\) and the remainder is \(x^{2}+3\): \[\frac {x^{3}+1}{x^{3}-x^{2}-2}=1+\frac {x^{2}+3}{x^{3}-x^{2}-2}.\]

\begin {align*} \text {Hence}\quad \frac {x^3 + 1}{\left (x^2 + 1\right ) \left ( x - 2\right )} & = 1 + \frac {x^2 + 3}{\left (x^2 + 1\right )\left (x - 2\right )}\\\\ & = 1 + \frac {A}{x - 2} + \frac {Bx + C}{x^2 + 1}\\\\ \end {align*}