Question

How do you evaluate the integral $\int{\dfrac{{{x}^{2}}}{{{\left( 4+{{x}^{2}} \right)}^{2}}}dx}$?

Answer 1

Hint: For solving the given integral we have to substitute $x=2\tan \theta $ so that in the denominator it will become $4\left( 1+{{\tan }^{2}}\theta \right)$ and we will apply the trigonometric $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $. Then we will obtain the integral in the form of \[{{\sin }^{2}}\theta \], which can be solved by using the identity $\cos 2\theta =1-2{{\sin }^{2}}\theta $. Finally, we will back substitute $\theta $ in the form of $x$ to get the final result.

Complete step-by-step answer:
Let us write the integral given in the question as
\[I=\int{\dfrac{{{x}^{2}}}{{{\left( 4+{{x}^{2}} \right)}^{2}}}dx}\]
Let us substitute
$x=2\tan \theta ........(i)$
Differentiating both sides
\[\begin{align}
  & \Rightarrow \dfrac{dx}{d\theta }=2{{\sec }^{2}}\theta \\
 & \Rightarrow dx=2{{\sec }^{2}}\theta d\theta ........(ii) \\
\end{align}\]
Substituting (i) and (ii) in the above integral, we get
\[\begin{align}
  & \Rightarrow I=\int{\dfrac{{{\left( 2\tan \theta \right)}^{2}}}{{{\left( 4+{{\left( 2\tan \theta \right)}^{2}} \right)}^{2}}}\left( 2{{\sec }^{2}}\theta d\theta \right)} \\
 & \Rightarrow I=\int{\dfrac{4{{\tan }^{2}}\theta \left( 2{{\sec }^{2}}\theta \right)}{{{\left( 4+4{{\tan }^{2}}\theta \right)}^{2}}}d\theta } \\
 & \Rightarrow I=\int{\dfrac{8{{\tan }^{2}}\theta {{\sec }^{2}}\theta }{{{4}^{2}}{{\left( 1+{{\tan }^{2}}\theta \right)}^{2}}}d\theta } \\
 & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{{{\tan }^{2}}\theta {{\sec }^{2}}\theta }{{{\left( 1+{{\tan }^{2}}\theta \right)}^{2}}}d\theta } \\
\end{align}\]
Now, we know the trigonometric identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $. Putting this above, we get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{{{\tan }^{2}}\theta {{\sec }^{2}}\theta }{{{\left( {{\sec }^{2}}\theta \right)}^{2}}}d\theta } \\
 & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{{{\tan }^{2}}\theta }{{{\sec }^{2}}\theta }d\theta } \\
\end{align}\]
Now, we put $\sec \theta =\dfrac{1}{\cos \theta }$ in the above integral to get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{{{\tan }^{2}}\theta }{\dfrac{1}{{{\cos }^{2}}\theta }}d\theta } \\
 & \Rightarrow I=\dfrac{1}{2}\int{{{\tan }^{2}}\theta {{\cos }^{2}}\theta d\theta } \\
\end{align}\]
Putting $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }{{\cos }^{2}}\theta d\theta } \\
 & \Rightarrow I=\dfrac{1}{2}\int{{{\sin }^{2}}\theta d\theta } \\
\end{align}\]
Now, we know that
$\begin{align}
  & \Rightarrow \cos 2\theta =1-2{{\sin }^{2}}\theta \\
 & \Rightarrow 2{{\sin }^{2}}\theta =1-\cos 2\theta \\
 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2} \\
\end{align}$
Putting this in the above integral, we get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\int{\left( \dfrac{1-\cos 2\theta }{2} \right)d\theta } \\
 & \Rightarrow I=\dfrac{1}{4}\int{\left( 1-\cos 2\theta \right)d\theta } \\
 & \Rightarrow I=\dfrac{1}{4}\left( \int{d\theta }-\int{\cos 2\theta d\theta } \right) \\
 & \Rightarrow I=\dfrac{1}{4}\left( \theta -\dfrac{\sin 2\theta }{2} \right)+C \\
\end{align}\]
Now, we know the trigonometric identity $\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }$. Putting this above, we get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{4}\left( \theta -\dfrac{2\tan \theta }{2\left( 1+{{\tan }^{2}}\theta \right)} \right)+C \\
 & \Rightarrow I=\dfrac{1}{4}\left( \theta -\dfrac{\tan \theta }{\left( 1+{{\tan }^{2}}\theta \right)} \right)+C.......(iii) \\
\end{align}\]
Now, from (i) we have
$\begin{align}
  & \Rightarrow x=2\tan \theta \\
 & \Rightarrow \tan \theta =\dfrac{x}{2}........(iv) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{x}{2} \right).......(v) \\
\end{align}$
Putting (iv) and (v) in (iii) we get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{4}\left( {{\tan }^{-1}}\left( \dfrac{x}{2} \right)-\dfrac{\dfrac{x}{2}}{\left( 1+{{\left( \dfrac{x}{2} \right)}^{2}} \right)} \right)+C. \\
 & \Rightarrow I=\dfrac{1}{4}\left( {{\tan }^{-1}}\left( \dfrac{x}{2} \right)-\dfrac{x}{2\left( 1+\dfrac{{{x}^{2}}}{4} \right)} \right)+C \\
 & \Rightarrow I=\dfrac{1}{4}\left( {{\tan }^{-1}}\left( \dfrac{x}{2} \right)-\dfrac{2x}{\left( {{x}^{2}}+4 \right)} \right)+C \\
\end{align}\]
Hence, this is the required integration.

Note: Do not forget to back substitute the variable $\theta $ in terms of $x$ in the final step, that is, $\tan \theta =\dfrac{x}{2}$. This is necessary since $\theta $ is our assumed variable, but the actual variable in the given equation is $x$. Also, we can solve this question by splitting the function inside the given integral, $\dfrac{{{x}^{2}}}{{{\left( 4+{{x}^{2}} \right)}^{2}}}$ into two fractions. For this, we need to add and subtract $4$ in the denominator to get $\dfrac{4+{{x}^{2}}-4}{{{\left( 4+{{x}^{2}} \right)}^{2}}}=\dfrac{1}{\left( 4+{{x}^{2}} \right)}-\dfrac{4}{{{\left( 4+{{x}^{2}} \right)}^{2}}}$. Lastly, since the given integral is indefinite, do not forget to add a constant after performing the integration.