Question

How do you find the integral of $\dfrac{{dx}}{{{{\left( {{x^2} - 4} \right)}^2}}}$

Answer 1

To solve this question first we substitute the value of x as $u = \dfrac{1}{2}x$. Then we differentiate that and put the value of $dx$ also. Then we substitute the value of $u$ in terms of hyperbolic trigonometry. And then we integrate that function and then we substitute again all the values and express the expression in terms of $x$ to get the final answer.

Complete step-by-step answer:
Let, the value of the integration be $i$.
$i = \int {\dfrac{{dx}}{{{{\left( {{x^2} - 4} \right)}^2}}}}$
To solve this question we will substitute $u = \dfrac{1}{2}x$
On differentiating both sides.
$du = \dfrac{1}{2}dx$
On squaring the substitution both sides.
${\left( u \right)^2} = {\left( {\dfrac{1}{2}x} \right)^2}$
On further solving
$4{u^2} = {x^2}$
On putting both these values in the first equation.
$i = \int {\dfrac{{2du}}{{{{\left( {4{u^2} - 4} \right)}^2}}}}$
Taking 4 outside the square and bracket
$i = \int {\dfrac{{2du}}{{16{{\left( {{u^2} - 1} \right)}^2}}}}$
Canceling the common factor from numerator and denominator
$i = \dfrac{1}{8}\int {\dfrac{{du}}{{{{\left( {{u^2} - 1} \right)}^2}}}}$
For further solving we have to again substitute the value of $u$.
$u = \tanh (t)$
Differentiating both side
Differentiation of $\dfrac{{d\tanh (t)}}{{dt}} = \dfrac{1}{{{{\cosh }^2}(t)}}$.
$du = \int {\dfrac{1}{{{{\cosh }^2}(t)}}dt}$
On putting both these values in the original equation.
$i = \int {\dfrac{{\dfrac{1}{{{{\cosh }^2}(t)}}dt}}{{8{{\left( {{{\tanh }^2}\left( t \right) - 1} \right)}^2}}}}$
If we take negative from square then there is no effect on expression.
$i = \int {\dfrac{{dt}}{{8{{\cosh }^2}(t){{\left( {1 - {{\tanh }^2}\left( t \right)} \right)}^2}}}}$
We know that $1 - {\tanh ^2}\left( t \right) = \dfrac{1}{{{{\cosh }^2}\left( t \right)}}$
On putting this value in the equation.
$i = \int {\dfrac{{dt}}{{8{{\cosh }^2}(t){{\left( {\dfrac{1}{{{{\cosh }^2}\left( t \right)}}} \right)}^2}}}}$
On solving this equation we get.
$i = \int {\dfrac{{{{\cosh }^2}(t)}}{8}dt}$
On making this equation in double angle formula ${\cosh ^2}(t) = \dfrac{1}{2}\left( {1 + \cosh (2t)} \right)$
$i = \dfrac{1}{{16}}\int {\left( {1 + \cosh (2t)} \right)dt}$
On integrating
$i = \dfrac{1}{{16}}\left( {t + \dfrac{1}{2}\sinh (2t)} \right) + c$
On again putting the value of $t$ in terms of $u$.
$i = \dfrac{1}{{16}}\left( {{{\tanh }^{ - 1}}(u) + \dfrac{1}{2}\sinh (2{{\tanh }^{ - 1}}(u))} \right) + c$
On putting the value of u in terms of x.
$i = \dfrac{1}{{16}}\left( {{{\tanh }^{ - 1}}(\dfrac{x}{2}) + \dfrac{1}{2}\sinh (2{{\tanh }^{ - 1}}(\dfrac{x}{2}))} \right) + c$
Final answer:
The value of integration of $\dfrac{{dx}}{{{{\left( {{x^2} - 4} \right)}^2}}}$is:
$i = \dfrac{1}{{16}}\left( {{{\tanh }^{ - 1}}(\dfrac{x}{2}) + \dfrac{1}{2}\sinh (2{{\tanh }^{ - 1}}(\dfrac{x}{2}))} \right) + c$

Note: To solve these types of questions there are many places where students often make mistakes. To solve these types of questions, students must have good practice of substitution and must know all the integrals and concepts of integral like trigonometry and hyperbolic trigonometry.