How do you find \[\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}}dx\] using partial fraction?

Hint: first we will convert an improper fraction into a proper fraction and then we will solve the question by a partial fraction. After that we will divide numerator and denominator by a long division method in which the quotient will be the whole number part, the remainder will be the numerator and the divisor will be treated as the denominator.

Complete step-by-step solution:
It is given that,
\[I=\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}}dx\].......eq(1)
 In this question, the degree of the highest term is two and the degree of numerator and denominator is equal. So, it is an improper fraction. First, we will divide eq(1) by the long division method, because we have to convert an improper fraction into a proper fraction. After doing a long division, we get
\[\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}dx=quotient+\dfrac{remainder}{divisor}}\]
On dividing by long division method, which is as shown below
\[{{x}^{2}}-4x+4\overset{3}{\overline{\left){\begin{align}
  & 3{{x}^{2}}\text{ }-10 \\
 & 3{{x}^{2}}-12x+12 \\
 & \overline{0\text{ }+12x-22} \\
\end{align}}\right.}}\]
We will divide this until we get the degree of remainder less than the divisor.
\[\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}dx=\int{3dx+\int{\dfrac{12x-22}{{{x}^{2}}-4x+4}}}}\]……eq(2)
Let, we will divide the integration into two parts \[{{I}_{1}}\]and \[{{I}_{2}}\]
\[\begin{align}
  & {{I}_{1}}=3 \\
 & {{I}_{2}}=\dfrac{12x-22}{{{x}^{2}}-4x+4} \\
 & I={{I}_{1}}+{{I}_{2}} \\
\end{align}\]
So, \[{{I}_{2}}=\int{\dfrac{12x-22}{{{x}^{2}}-4x+4}}\]
Now we will solve the \[{{I}_{2}}\] part with the help of a suitable identity first and then we will put \[{{I}_{2}}\] in eq(2)
\[\dfrac{12x-22}{\begin{align}
  & {{x}^{2}}-4x+4 \\
 & \\
\end{align}}=\dfrac{12x-22}{{{(x-2)}^{2}}}\]………..eq(3)
We have converted eq(1) into improper fraction from proper fraction, so now we will solve eq(3)
We know that, according to the identity of integration
\[\dfrac{Ax+B}{{{(x-a)}^{2}}}=\dfrac{A}{(x-a)}+\dfrac{B}{{{(x-a)}^{2}}}\]
\[\Rightarrow \dfrac{12x-22}{{{(x-2)}^{2}}}=\dfrac{A}{(x-2)}+\dfrac{B}{{{(x-2)}^{2}}}\]………..eq(4)
On taking the LCM and solving eq(4), we get
\[\dfrac{12x-22}{{{(x-2)}^{2}}}=\dfrac{A(x-2)+B}{{{(x-2)}^{2}}}\]
\[\begin{align}
  & \Rightarrow 12x-22=A(x-2)+B \\
 & \Rightarrow 12x-22=Ax-2A+B \\
\end{align}\]
On comparing R.H.S and L.H.S, we get the value of A and B and after that, we will put the values of A and B in eq(4)
\[A=12\]
\[-22=-2A+B\]…….eq(5)
On putting the value of A in eq(5), we get
\[\begin{align}
  & -22=-2(12)+B \\
 & \Rightarrow -22=-24+B \\
 & \Rightarrow -22+24=B \\
 & B=2 \\
\end{align}\]
On putting the value of A and B in eq(4), we get
\[\dfrac{12x-22}{{{(x-2)}^{2}}}=\dfrac{12}{(x-2)}+\dfrac{2}{{{(x-2)}^{2}}}\]
Value of \[{{I}_{2}}\] is now obtained, so now we will solve it further using a suitable formula of integration
Now putting this in eq(2), we get
\[\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}dx=\int{3dx+\int{\dfrac{12}{(x-2)}dx+\int{\dfrac{2}{{{(x-2)}^{2}}}}}}}\]…….eq(6)
\[\int{\dfrac{12}{x-2}dx=12\log \left| x-2 \right|}\]……..eq(7)
(using the formula of integration \[\int{\dfrac{1}{x}dx=\log x}\])
\[\int{\dfrac{2}{{{(x-2)}^{2}}}dx=2\int{{{(x-2)}^{-2}}}}\]
(using the identity of integration \[\int{{{x}^{-a}}=\int{\dfrac{{{x}^{-a+1}}}{-a+1}}}\])
\[\Rightarrow 2\int{{{(x-2)}^{-2}}=}2\int{\dfrac{{{(x-2)}^{-2+1}}}{-2+1}}\]
\[\int{\dfrac{2}{{{(x-2)}^{2}}}dx=-2\int{\dfrac{1}{x-2}dx}}\]
\[\int{\dfrac{2}{{{(x-2)}^{2}}}dx=-2\log \left| x-2 \right|}\]……..eq(8)
Putting eq(7) and eq(8) in eq(6), we get
\[\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}=\int{3dx+\int{\dfrac{12}{(x-2)}dx+\int{\dfrac{2}{{{(x-2)}^{2}}}dx}}}}\]
\[\int{\dfrac{3{{x}^{2}}-10}{{{x}^{2}}-4x+4}=3x+12\log \left| x-2 \right|-2\log \left| x-2 \right|}\]
Which is the required answer

Note: If the degree of the numerator is less than the degree of the denominator then it is known as a proper fraction. If the degree of the numerator is greater than the degree of the denominator or if the degree of both numerator and denominator is the same then it is known as improper fraction.
\[{{x}^{2}}-4x+4\overset{3}{\overline{\left){\begin{align}
  & 3{{x}^{2}}\text{ }-10 \\
 & 3{{x}^{2}}-12x+12 \\
 & \overline{0\text{ }+12x-22} \\
\end{align}}\right.}}\]