JEE Main 2021 (Online) 18th March Morning ShiftQuestion 1
If $f(x)=\int\frac{5x^{8}+7x^{6}}{\left(x^{2}+1+2x^{7}\right)^{2}}dx$, $(x\ge0)$, $f(0)=0$ and $f(1)=\frac{1}{K}$, then the value of $K$ is $\_\_\_\_\_\_$ .
Indefinite Integration
Answer$4$
Solution
Concept. Divide top and bottom by $x^{14}$; the numerator becomes $-\frac{d}{dx}\big(x^{-5}+x^{-7}\big)$. Formulas used. (1) $\dfrac{5x^8+7x^6}{(x^2+1+2x^7)^2}=\dfrac{5x^{-6}+7x^{-8}}{\big(x^{-5}+x^{-7}+2\big)^2}$(2) $w=x^{-5}+x^{-7}$: $dw=-\big(5x^{-6}+7x^{-8}\big)dx$Step 1 - Substitute. By (1), (2), $$f=-\int\frac{dw}{(w+2)^2}=\frac1{w+2}+C=\frac{x^7}{x^2+1+2x^7}+C.$$ Step 2 - Constant. $f(0)=0$ gives $C=0$. Step 3 - Value. $f(1)=\frac14$, so $K=4$. Answer: 4