If f(x) = \int 5x^8 + 7x^6 \over (x^2 + 1 + 2x^7)^2dx,(x \ge 0),f(0) = 0 and f(1) = 1 \over K , then the value of K is

Indefinite Integrals PYQ — JEE Main 2021 · 18 Mar · Shift 1 · Q45 with Solution

View written solutionFree

Correct answer: 4

We need to evaluate $f(x)=\int \frac{5x^8+7x^6}{(x^2+1+2x^7)^2}\,dx,$ with the condition $f(0)=0$ , and then use $f(1)=\frac1K$ .
Observe the denominator: $x^2+1+2x^7 = 1+x^2+2x^7.$ Let $u=1+x^2+2x^7.$ Then $\frac{du}{dx}=2x+14x^6=2x(1+7x^5).$ This does not immediately match the numerator, so let us inspect the integrand more carefully.
Factor the numerator: $5x^8+7x^6=x^6(5x^2+7).$ Now rewrite the denominator as $(x^7+x^5)^2 \text{?}$ That does not help. So instead, check whether the integrand is a derivative of a simple quotient.

Try $\frac{d}{dx}\left(\frac{x^7}{1+x^2+2x^7}\right).$ Using quotient rule,

=\frac{7x^6(1+x^2+2x^7)-x^7(2x+14x^6)}{(1+x^2+2x^7)^2}.$$ Simplify the numerator: $$7x^6+7x^8+14x^{13}-2x^8-14x^{13}=7x^6+5x^8.$$ So, $$\frac{d}{dx}\left(\frac{x^7}{1+x^2+2x^7}\right)=\frac{5x^8+7x^6}{(1+x^2+2x^7)^2}.$$ This matches the integrand exactly. 4. Therefore, $$f(x)=\frac{x^7}{1+x^2+2x^7}+C.$$ Using $f(0)=0$: $$f(0)=\frac{0}{1+0+0}+C=0 \implies C=0.$$ Hence, $$f(x)=\frac{x^7}{1+x^2+2x^7}.$$ 5. Now compute $f(1)$: $$f(1)=\frac{1^7}{1+1^2+2\cdot 1^7}= rac{1}{1+1+2}= rac14.$$ Since $$f(1)=\frac1K,$$ we get $$\frac1K=\frac14 \implies K=4.$$ 6. Final answer: $$\boxed{4}$$ The derived answer matches the stored correct answer.