Question

Evaluate: $\int{\dfrac{\sqrt{{{x}^{2}}-8}}{{{x}^{4}}}dx}$

Answer 1

Hint: Express ${{x}^{4}}$ as multiplication of $x$ and ${{x}^{3}}$and take $'x'$ to the square root term written in numerator of the given expression. And Substitute the whole term inside the square root as ${{t}^{2}}$ . Now differentiate both the sides and hence, solve it. Use the following formula:- $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$.

Complete step-by-step answer:
Let the value of the given integral be I. So, we can write an equation as
$I=\int{\dfrac{\sqrt{{{x}^{2}}-8}}{{{x}^{4}}}}dx$ $.......(i)$
As such, we can write the term ${{x}^{4}}$ in the denominator of the above expression as $x.{{x}^{3}}$. So, we can get equation (i) as
$I=\int{\dfrac{\sqrt{{{x}^{2}}-8}}{x.{{x}^{3}}}dx.......(ii)}$
Now, we can send in ‘x’ the denominator to the square root term in the numerator i.e. inside the term $\sqrt{{{x}^{2}}-8}$ . Hence, the term in denominator will become inside the square root to balance the equation. So, we get I as
$I=\int{\dfrac{\sqrt{\dfrac{{{x}^{2}}-8}{{{x}^{2}}}}}{{{x}^{3}}}dx}$
$orI=\int{\dfrac{1}{{{x}^{3}}}}\sqrt{1-\dfrac{8}{{{x}^{2}}}}dx.......(iii)$
As, we can write the term in the denominator of the above expression as well. So, we can get equation (i) as
Now, we can send ‘x’ in the denominator to the square root term in the numerator i.e. inside the term . Hence, the term ‘x’ in the denominator will become inside the square root to balance the equation. So, we get ‘I’ as
Now, we can suppose the whole root part as and get the equation as

$1-\dfrac{8}{{{x}^{2}}}={{t}^{2}}$ $........(iv)$
Differentiating the whole expression w.r.t ‘x’, we get
$\dfrac{d}{dx}\left( 1-\dfrac{8}{{{x}^{2}}} \right)=\dfrac{d}{dx}\left( {{t}^{2}} \right).........(v)$
Now, we know the differentiation of ${{x}^{n}}$ and constant can be given as
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\dfrac{d}{dx}(cons\tan t)=0$
So, we can get equation (v) by using the above relation, as
$\dfrac{d}{dx}(1)-8\dfrac{d}{dx}{{x}^{-2}}=\dfrac{d}{dx}{{t}^{2}}$

& 0-8(-2){{x}^{-2-1}}=2t\dfrac{dt}{dx} \\\ & 16{{x}^{-3}}=2t\dfrac{dt}{dx} \\\ & or\dfrac{8}{{{x}^{3}}}=t\dfrac{dt}{dx} \\\ \end{aligned}$$ On Cross multiplying the above equation, we get value of $\dfrac{dx}{{{x}^{3}}}$ as $\dfrac{dx}{{{x}^{3}}}=\dfrac{t}{8}dt.........(vi)$ Now, we can put the value of $\dfrac{dx}{{{x}^{3}}}$ as $\dfrac{t}{8}dt$ from the equation (vi) as and value of $$1-\dfrac{8}{{{x}^{2}}}$$as ${{t}^{2}}$ from the equation (iv), in the equation (iii), so we get $I=\int{\dfrac{t}{8}\times {{({{t}^{2}})}^{1/2}}dt}$ $I=\int{\dfrac{t}{8}tdt=\int{\dfrac{{{t}^{2}}}{8}dt}}$ $I=\dfrac{1}{8}\int{{{t}^{2}}dt}$ As, we know the integration of ${{x}^{n}}$ can be given as $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ So, value of I can be given as $\begin{aligned} & I=\dfrac{1}{8}\int{{{t}^{2}}dt} \\\ & I=\dfrac{1}{8}\dfrac{{{t}^{3}}}{3}+c \\\ & I=\dfrac{{{t}^{3}}}{24}+c.......(vii) \\\ \end{aligned}$ Value of t as $\begin{aligned} & {{t}^{2=}}1-\dfrac{8}{{{x}^{2}}} \\\ & t=\sqrt{1-\dfrac{8}{{{x}^{2}}}}={{\left( 1-\dfrac{8}{{{x}^{2}}} \right)}^{1/2}} \\\ \end{aligned}$ So, we can get the equation (vii) in terms of x as $$\begin{aligned} & I=\dfrac{1}{24}{{\left( 1-\dfrac{8}{{{x}^{2}}} \right)}^{3/2}}+c \\\ & I=\dfrac{1}{24}\left( 1-\dfrac{8}{{{x}^{2}}} \right)\sqrt{1-\dfrac{8}{{{x}^{2}}}}+c \\\ \end{aligned}$$ Hence, value of given integral i.e. $$\int{\dfrac{\sqrt{{{x}^{2}}-8}}{{{x}^{4}}}dx}$$ is $\dfrac{1}{24}\left( 1-\dfrac{8}{{{x}^{2}}} \right)\sqrt{1-\dfrac{8}{{{x}^{2}}}}+c$ Note: Another approach for the given question would be that we can multiply and divide the expression in root (numerator) by ${{x}^{2}}$ as $\begin{aligned} & I=\int{\dfrac{\sqrt{\dfrac{{{x}^{2}}-8}{{{x}^{2}}}}\times {{x}^{2}}}{{{x}^{4}}}dx} \\\ & I=\int{\dfrac{x}{{{x}^{4}}}\sqrt{\dfrac{{{x}^{2}}-8}{{{x}^{2}}}}dx=\int{\dfrac{1}{{{x}^{3}}}\sqrt{1-\dfrac{8}{{{x}^{2}}}}dx}} \\\ \end{aligned}$ $$$$ Now, proceed with the same approach as done in the solution. Converting the given term by the observation to the other, for applying the substitution method is the key point of the question. For applying the substitution in any integral, any term with its derivative should be present there.