Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given integral is

[tex]\rm \: \displaystyle\int\rm \: \frac{ {x}^{3} + 3 }{ {x}^{2} - 7x + 12} dx\\ \\ [/tex]

Since, degree of numerator is greater than degree of denominator, so we have to first divide numerator by denominator.

On apply Long Division Method, we have

[tex]\rm \: = \: \displaystyle\int\rm \:\bigg(x + 7 + \frac{37x - 81}{ {x}^{2} - 7x + 12} \bigg)dx\\ \\ [/tex]

[tex]\rm \: = \: \dfrac{ {x}^{2} }{2} + 7x + I - - - - (1) \\ \\ [/tex]

where,

[tex]\rm \: I= \: \displaystyle\int\rm \: \frac{37x - 81}{ {x}^{2} - 7x + 12}dx\\ \\ [/tex]

can be further rewritten as

[tex]\rm \: I= \: \displaystyle\int\rm \: \frac{\dfrac{37}{2}(2x) - 81}{ {x}^{2} - 7x + 12}dx\\ \\ [/tex]

[tex]\rm \: I= \: \displaystyle\int\rm \: \frac{\dfrac{37}{2}(2x - 7 + 7) - 81}{ {x}^{2} - 7x + 12}dx\\ \\ [/tex]

[tex]\rm \: I= \: \displaystyle\int\rm \: \frac{\dfrac{37}{2}(2x - 7) + \dfrac{259}{2} - 81}{ {x}^{2} - 7x + 12}dx\\ \\ [/tex]

[tex]\rm \: I= \: \displaystyle\int\rm \: \frac{\dfrac{37}{2}(2x - 7) + \dfrac{259 - 162}{2} }{ {x}^{2} - 7x + 12}dx\\ \\ [/tex]

[tex]\rm \: I= \: \displaystyle\int\rm \: \frac{\dfrac{37}{2}(2x - 7) + \dfrac{97}{2} }{ {x}^{2} - 7x + 12}dx\\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} \displaystyle\int\rm \: \frac{(2x - 7)}{ {x}^{2} - 7x + 12}dx + \dfrac{97}{2}\displaystyle\int\rm \: \frac{dx}{ {x}^{2} - 7x + 12 } \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log | {x}^{2} - 7x + 12 | + \dfrac{97}{2}\displaystyle\int\rm \: \frac{dx}{ {x}^{2} - 7x + \dfrac{49}{4} - \dfrac{49}{4} + 12 } \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log | {x}^{2} - 7x + 12 | + \dfrac{97}{2}\displaystyle\int\rm \: \frac{dx}{ {\bigg(x - \dfrac{7}{2} \bigg) }^{2} + \dfrac{48 - 49}{4} } \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log | {x}^{2} - 7x + 12 | + \dfrac{97}{2}\displaystyle\int\rm \: \frac{dx}{ {\bigg(x - \dfrac{7}{2} \bigg) }^{2} - \dfrac{1}{4} } \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log | {x}^{2} - 7x + 12 | + \dfrac{97}{2}\displaystyle\int\rm \: \frac{dx}{ {\bigg(x - \dfrac{7}{2} \bigg) }^{2} - {\bigg(\dfrac{1}{2} \bigg) }^{2} } \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log | {x}^{2} - 7x + 12 | + \dfrac{97}{2} \times \dfrac{1}{2 \times \dfrac{1}{2}}log\bigg |\dfrac{x - \dfrac{7}{2} - \dfrac{1}{2}}{x - \dfrac{7}{2} + \dfrac{1}{2}} \bigg| + c \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log | {x}^{2} - 7x + 12 | + \dfrac{97}{2} log\bigg |\dfrac{x - 4}{x -3} \bigg| + c \\ \\ [/tex]

[tex]\rm \: I= \: \dfrac{37}{2} log |(x - 4)(x - 3)| + \dfrac{97}{2} log\bigg |\dfrac{x - 4}{x -3} \bigg| + c \\ \\ [/tex]

[tex]\rm \: = \dfrac{37}{2} log |x - 4| + \dfrac{37}{2}log |x - 3| + \dfrac{97}{2} log |x - 4| - \dfrac{97}{2}log |x - 3| + c \\ \\ [/tex]

[tex]\rm \: =67 \: log |x - 4| - 30 \: log |x - 3| + c \\ \\ [/tex]

On substituting this value in equation (1), we get

[tex]\rm \: = \: \frac{ {x}^{2} }{2} + 7x + 67 \: log |x - 4| - 30 \: log |x - 3| + c \\ \\ [/tex]

Hence,

[tex]\bf\implies \:\int \:\frac{ {x}^{3} + 3 }{ {x}^{2} - 7x + 12}dx \\ \\ \rm \: = \: \frac{ {x}^{2} }{2} + 7x + 67 \: log |x - 4| - 30 \: log |x - 3| + c \\ \\ [/tex]