NIMCET 2022 — Mathematics PYQ

Step-by-step Solution 
Step 1: Rewrite the integrand 
The given integral is $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$. The term inside the square root can be rewritten by dividing by $x^4$: $\sqrt{2x^4-2x^2+1}=\sqrt{x^4(2-\frac{2}{x^2}+\frac{1}{x^4})}=x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}$. Substitute this back into the integral: $\int \frac{x^2-1}{x^3\cdot x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx=\int \frac{x^2-1}{x^5\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$. Divide the numerator and denominator by $x^3$: $\int \frac{\frac{x^2-1}{x^3}}{\frac{x^5}{x^3}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx=\int \frac{\frac{1}{x}-\frac{1}{x^3}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$. [1, 2]  
Step 2: Perform a substitution 
Let $t=2-\frac{2}{x^2}+\frac{1}{x^4}$. Differentiate $t$ with respect to $x$: $\frac{dt}{dx}=\frac{d}{dx}(2-2x^{-2}+x^{-4})=0-2(-2)x^{-3}+(-4)x^{-5}=\frac{4}{x^3}-\frac{4}{x^5}=4(\frac{1}{x^3}-\frac{1}{x^5})$. This can be rewritten as $dt=4(\frac{1}{x^3}-\frac{1}{x^5})dx$. Notice that the numerator of the integral is $\frac{1}{x}-\frac{1}{x^3}$. This is not directly $dt$. Let's re-evaluate the substitution. Consider $u=\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}$. Then $u^2=2-\frac{2}{x^2}+\frac{1}{x^4}$. Differentiate $u^2$ with respect to $x$: $2u\frac{du}{dx}=\frac{4}{x^3}-\frac{4}{x^5}=4(\frac{1}{x^3}-\frac{1}{x^5})$. So, $u\frac{du}{dx}=2(\frac{1}{x^3}-\frac{1}{x^5})$. This still does not match the numerator. 
Let's try a different substitution based on the structure of the integrand. Let $y=\frac{1}{x}$. Then $dy=-\frac{1}{x^2}dx$. The integral can be written as $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int \frac{1-\frac{1}{x^2}}{x\sqrt{2x^4-2x^2+1}}dx$. This is not simplifying well. [1, 2]  
Let's go back to the expression $\int \frac{\frac{1}{x}-\frac{1}{x^3}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$. Let $v=2-\frac{2}{x^2}+\frac{1}{x^4}$. Then $dv=(\frac{4}{x^3}-\frac{4}{x^5})dx=4(\frac{1}{x^3}-\frac{1}{x^5})dx$. The numerator is $\frac{1}{x}-\frac{1}{x^3}$. This is not directly related to $dv$. 
Let's try a substitution for the term inside the square root. Let $u=2x^4-2x^2+1$. Then $du=(8x^3-4x)dx=4x(2x^2-1)dx$. This does not match the numerator. 
Let's reconsider the first step and simplify the integrand differently. $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int \frac{x^2-1}{x^3\cdot x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx=\int \frac{x^2-1}{x^5\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$. Divide the numerator and denominator by $x^5$: $\int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$. Let $t=2-\frac{2}{x^2}+\frac{1}{x^4}$. Then $dt=(\frac{4}{x^3}-\frac{4}{x^5})dx=4(\frac{1}{x^3}-\frac{1}{x^5})dx$. So, $\frac{1}{4}dt=(\frac{1}{x^3}-\frac{1}{x^5})dx$. The integral becomes $\int \frac{1}{\sqrt{t}}\cdot \frac{1}{4}dt=\frac{1}{4}\int t^{-1/2}dt$. [2, 3]  
Step 3: Integrate with respect to $t$ 
$\frac{1}{4}\int t^{-1/2}dt=\frac{1}{4}\frac{t^{1/2}}{1/2}+C=\frac{1}{4}\cdot 2\sqrt{t}+C=\frac{1}{2}\sqrt{t}+C$. 
Step 4: Substitute back for $t$ 
Substitute $t=2-\frac{2}{x^2}+\frac{1}{x^4}$ back into the expression: $\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$. [2]  
Final Answer 
The final answer is $\boxed{\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C}$.