JEE Main 2019 (Online) 9th January Evening ShiftQuestion 1
If $f(x)=\int\frac{5x^{8}+7x^{6}}{\left(x^{2}+1+2x^{7}\right)^{2}}dx$, $(x\ge0)$, and $f(0)=0$, then the value of $f(1)$ is
Indefinite Integration
Solution
Concept. Divide by $x^{14}$; the numerator becomes minus the derivative of $2+x^{-5}+x^{-7}$. Formulas used. (1) $t=2+x^{-5}+x^{-7}\Rightarrow dt=-\left(5x^{-6}+7x^{-8}\right)dx$Step 1 - Rewrite. $$\frac{5x^8+7x^6}{(x^2+1+2x^7)^2}=\frac{5x^{-6}+7x^{-8}}{(2+x^{-5}+x^{-7})^2}.$$ Step 2 - Use (1). $$-\int\frac{dt}{t^2}=\frac1t+C=\frac{x^7}{2x^7+x^2+1}+C.$$ $f(0)=0$ gives $C=0$. Step 3 - Value. $f(1)=\dfrac{1}{2+1+1}=\dfrac14$. Answer: (C)