JEE Main 2014 — Mathematics Calculus

$$\int \frac{{x}^{5m-1}+2{x}^{4m-1}}{{({x}^{2m}+{x}^{m}+1)}^{3}}dx=f(x)+c$$∫(x2m+xm+1)3x5m−1+2x4m−1​dx=f(x)+c

$$\Rightarrow \int \frac{{x}^{5m-1}+2{x}^{4m-1}}{{x}^{6m}{(1+{x}^{-m}+{x}^{-2m})}^{3}}dx=f(x)+c$$⇒∫x6m(1+x−m+x−2m)3x5m−1+2x4m−1​dx=f(x)+c

$\displaystyle \Rightarrow \int \frac{{x}^{-m-1}+2{x}^{-2m-1}}{{(1+{x}^{-m}+{x}^{-2m})}^{3}}dx=f(x)+c$⇒∫(1+x−m+x−2m)3x−m−1+2x−2m−1​dx=f(x)+c.

If, $1+{x}^{-m}+{x}^{-2m}=t$1+x−m+x−2m=t

$$\Rightarrow -m({x}^{-m-1}+2{x}^{-2m-1})dx=dt$$⇒−m(x−m−1+2x−2m−1)dx=dt

$$\therefore \int \frac{-dt}{m{t}^{3}}=f(x)+c$$∴∫mt3−dt​=f(x)+c

$$\Rightarrow -\frac{1}{m}\int {t}^{-3}dt=f(x)+c$$⇒−m1​∫t−3dt=f(x)+c

Using $\displaystyle \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C,n\neq -1$∫xndx=n+1xn+1​+C,n=−1

$$\Rightarrow -\frac{1}{m}(\frac{{t}^{-2}}{-2})=f(x)+c$$⇒−m1​(−2t−2​)=f(x)+c

$$\Rightarrow \frac{1}{2m{(1+{x}^{-m}+{x}^{-2m})}^{2}}+c=f(x)+c$$⇒2m(1+x−m+x−2m)21​+c=f(x)+c

$$\Rightarrow \frac{1}{2m{(1+\frac{1}{{x}^{m}}+\frac{1}{{x}^{2m}})}^{2}}+c=f(x)+c$$⇒2m(1+xm1​+x2m1​)21​+c=f(x)+c

$$\Rightarrow \frac{1}{2m}\frac{{x}^{4m}}{{({x}^{2m}+{x}^{m}+1)}^{2}}=f(x).$$⇒2m1​(x2m+xm+1)2x4m​=f(x).