Question

If $\int{\dfrac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}}=A\left( {{\tan }^{-1}}\left( \dfrac{x-1}{3} \right)+\dfrac{f\left( x \right)}{{{x}^{2}}-2x+10} \right)+C$ Where C is the integration constant. Then which of the following is true.
$\begin{align}
  & a)A=\dfrac{1}{27}\text{ and f}\left( x \right)=9\left( x-1 \right) \\
 & b)A=\dfrac{1}{81}\text{ and f}\left( x \right)=3\left( x-1 \right) \\
 & c)A=\dfrac{1}{54}\text{ and f}\left( x \right)=9{{\left( x-1 \right)}^{2}} \\
 & d)A=\dfrac{1}{54}\text{ and f}\left( x \right)=3\left( x-1 \right) \\
\end{align}$

Answer 1

Hint: Now we have the given equation in form of $\int{\dfrac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}}$ and we will try to write it in the form of $I=\int{\dfrac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+9 \right)}^{2}}}}$ . Now we will substitute x – 1 as 3tant and use the identity ${{\tan }^{2}}t+1={{\sec }^{2}}t$. Using this substitution we will get a simplified integral. We will further simplify it more with formula \[\cos 2x=2{{\cos }^{2}}x-1\] and hence solve the integral.

Complete step by step answer:
Now first let us consider $I=\int{\dfrac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}}$
We will first try to write it in the form of $\int{\dfrac{dx}{{{\left( {{\left( x-a \right)}^{2}}+b \right)}^{2}}}}$
Hence we write 10 as 9 + 1 and use 1 to complete the square
$\int{\dfrac{dx}{{{\left( {{x}^{2}}-2x+1+9 \right)}^{2}}}}$
Now we know that $\left( {{a}^{2}}-2ab+{{b}^{2}} \right)={{\left( a-b \right)}^{2}}$ using this we get
$I=\int{\dfrac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+9 \right)}^{2}}}}$
Now let us substitute $\left( x-1 \right)=3\tan t$
Then we know that $dx=3{{\sec }^{2}}tdt$
Substituting this we get.
\[I=\int{\dfrac{se{{c}^{2}}tdt}{{{\left( {{\left( 3{{\tan }^{2}}t \right)}^{2}}+9 \right)}^{2}}}}\]
\[\begin{align}
  & I=\int{\dfrac{{{\sec }^{2}}tdt}{{{\left( 9{{\tan }^{2}}t+9 \right)}^{2}}}} \\
 & I=\int{\dfrac{1}{27}\dfrac{{{\sec }^{2}}tdt}{{{\left( {{\tan }^{2}}t+1 \right)}^{2}}}} \\
 & I=\dfrac{1}{27}\int{\dfrac{{{\sec }^{2}}tdt}{{{\left( {{\tan }^{2}}t+1 \right)}^{2}}}} \\
\end{align}\]
Now we know the trigonometric identity that ${{\tan }^{2}}t+1={{\sec }^{2}}t$
Hence we get
$\begin{align}
  & I=\dfrac{1}{27}\int{\dfrac{se{{c}^{2}}tdt}{{{\left( {{\sec }^{2}}t \right)}^{2}}}} \\
 & I=\dfrac{1}{27}\int{\dfrac{dt}{{{\sec }^{2}}t}} \\
\end{align}$
Now we know that $\sec x$ is nothing but $\dfrac{1}{\cos x}$
Hence using this we get the value of I as
\[I=\dfrac{1}{27}\int{{{\cos }^{2}}tdt}\]
Now we know that \[\cos 2x=2{{\cos }^{2}}x-1\Rightarrow {{\cos }^{2}}x=\dfrac{\cos 2x+1}{2}\]
Hence using this substitution we get
\[\begin{align}
  & I=\dfrac{1}{27}\int{\left( \dfrac{\cos 2t+1}{2} \right)} \\
 & I=\dfrac{1}{27}\int{\dfrac{\cos 2tdt}{2}}+\dfrac{1}{27}\int{\dfrac{dt}{2}} \\
 & I=\dfrac{1}{54}\int{\cos 2tdt}+\dfrac{1}{54}\int{dt} \\
\end{align}\]
Now we know that integration of $\cos 2tdt$ is nothing but $\dfrac{\sin 2t}{2}$ and integration of dt is t
Hence we get
$I=\dfrac{1}{54}\left( \dfrac{\sin 2t}{2} \right)+\dfrac{t}{54}+C......................\left( 1 \right)$
Now consider $\dfrac{\sin 2t}{2}$
Using sin2x = 2sinx cosx we get
$\dfrac{\sin 2t}{2}=\dfrac{2\sin t\cos t}{2}$
Now dividing the whole equation by ${{\cos }^{2}}t$
$\begin{align}
  & =\dfrac{\dfrac{\sin t\cos t}{{{\cos }^{2}}t}}{\dfrac{1}{{{\cos }^{2}}t}} \\
 & =\dfrac{\tan t}{{{\sec }^{2}}t} \\
\end{align}$
Now we know that ${{\tan }^{2}}t+1={{\sec }^{2}}t$ hence we have
$\dfrac{\sin 2t}{t}=\dfrac{\tan t}{1+{{\tan }^{2}}t}$
Now substituting this in equation (1) we get
$\begin{align}
  & I=\dfrac{1}{54}\left( \dfrac{\tan t}{1+{{\tan }^{2}}t} \right)+\dfrac{t}{54}+C \\
 & I=\dfrac{1}{54}\left( \dfrac{\tan t}{1+{{\tan }^{2}}t} \right)+\dfrac{t}{54}+C \\
\end{align}$
Now if we resubstitute the value of t we get which is $t={{\tan }^{-1}}\dfrac{\left( x-1 \right)}{3}$ we get
$I=\dfrac{1}{54}\left( \left( \dfrac{\dfrac{x-1}{3}}{1+{{\left( \dfrac{x-1}{3} \right)}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{x-1}{3} \right) \right)+C$
$\begin{align}
  & I=\dfrac{1}{54}\left( \dfrac{\dfrac{x-1}{3}}{\dfrac{9+{{x}^{2}}-2x+1}{9}} \right)+{{\tan }^{-1}}\left( \dfrac{x-1}{3} \right)+C \\
 & I=\dfrac{1}{54}\left( \dfrac{x-1}{\dfrac{9+{{x}^{2}}-2x+1}{3}} \right)+{{\tan }^{-1}}\left( \dfrac{x-1}{3} \right)+C \\
 & I=\dfrac{1}{54}\left( \dfrac{3\left( x-1 \right)}{9+{{x}^{2}}-2x+1} \right)+{{\tan }^{-1}}\left( \dfrac{x-1}{3} \right)+C \\
\end{align}$
Hence comparing the above equation with $A\left( {{\tan }^{-1}}\left( \dfrac{x-1}{3} \right)+\dfrac{f\left( x \right)}{{{x}^{2}}-2x+10} \right)+C$ we get
$A=\dfrac{1}{54},f\left( x \right)=3\left( x-1 \right)$
Option b is the correct option.

Note:
Now whenever we arrive at terms like ${{\left( x\pm a \right)}^{n}},{{\left( a\pm x \right)}^{n}}$ we can think of solving the integration by substitution of trigonometric ratio. The trigonometric ratio though must be chosen wisely such that any one of the trigonometric identities can be applied.
For example if we have $\left( {{a}^{2}}+{{x}^{2}} \right)$ then we know that by substituting x = atant we will get ${{a}^{2}}\left( 1+{{\tan }^{2}}t \right)$ hence we can use $1+{{\tan }^{2}}t={{\sec }^{2}}t$.