Question

Evaluate the given integral:
$\int{\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}dx}$.

Answer 1

We can clearly observe that in the given problem ${{x}^{2}}$ is repeating 3 times one time in numerator and 2 times at denominator so we can substitute the ${{x}^{2}}$ with some random variable say t, then we will convert the given integral in terms of t. and then we will apply the method of partial fractions to solve the obtained integral.

Complete step by step answer:
To solve the given integral,
$\int{\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}dx}$
We can see that the term ${{x}^{2}}$ in repeating 3 times in the integral so we will change the term ${{x}^{2}}$ with a random variable i.e. t to make it more simple, so we get
${{x}^{2}}=t$
Then the term $\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}$ can also be represented as $\dfrac{t}{\left( t+4 \right)\left( t+9 \right)}$,
Now we will try to solve the given integral by partial fraction method, so we get
$\dfrac{t}{\left( t+4 \right)\left( t+9 \right)}=\dfrac{A}{\left( t+4 \right)}+\dfrac{B}{\left( t+9 \right)}$
Taking the LCM of the RHS of the equation,
$\dfrac{t}{\left( t+4 \right)\left( t+9 \right)}=\dfrac{A\left( t+9 \right)+B\left( t+4 \right)}{\left( t+4 \right)\left( t+9 \right)}$
Now, comparing the LHS and RHS we get
$t=A\left( t+9 \right)+B\left( t+4 \right)$
Rearranging the above expression, we get
$t=\left( A+B \right)t+\left( 9A+4B \right)$
Now again comparing both sides we have,
$A+B=1$
And,
$9A+4B=0$
Solving the above two equations, we get
$A=-\dfrac{4}{5}\,and\,B=\dfrac{9}{5}$
Hence,
$\Rightarrow \dfrac{t}{\left( t+4 \right)\left( t+9 \right)}=-\dfrac{4}{5\left( t+4 \right)}+\dfrac{9}{5\left( t+9 \right)}$
Putting the value of t back in the above equation, we get
$\int{\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}dx=\int{-\dfrac{4}{5\left( {{x}^{2}}+4 \right)}dx+\int{\dfrac{9}{5\left( {{x}^{2}}+9 \right)}dx}}}$
$\int{\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}dx}=-\dfrac{4}{5}\int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx}+\dfrac{9}{5}\int{\dfrac{1}{\left( {{x}^{2}}+9 \right)}}dx$
Now we know that $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}}\,+\,C$ so applying this in above equation, we get
$\int{\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}dx}=-\dfrac{4}{5}\times \dfrac{1}{2}{{\tan }^{-1}}\dfrac{x}{2}+\dfrac{9}{5}\times \dfrac{1}{3}{{\tan }^{-1}}\dfrac{x}{3}+C$

Hence, $\int{\dfrac{{{x}^{2}}}{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+9 \right)}dx}=-\dfrac{2}{5}{{\tan }^{-1}}\dfrac{x}{2}+\dfrac{3}{5}{{\tan }^{-1}}\dfrac{x}{3}+C$

Note: You may make this mistake while solving this that instead of converting it into partial fraction form you may substitute it into sine or cosine form but by that method you will not be able to solve it easily rather you will get stuck in complex equation so follow the method shown in the solution and try to analyse which steps we performed and why. And also remember the integration $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}}\,+\,C$ for future references.