If \(f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0), f(0)=0\) and \(f(1)=\frac{1}{K}\), then the value of \(K\) is

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Indefinite Integrals

f (x) = \int \frac{5 x ^{8} + 7 x ^{6}}{( x ^{2} + 1 + 2 x ^{7} ) ^{2}} d x, (x \geq 0), f (0) = 0

and

f (1) = \frac{1}{K}

, then the value of

K

Answer: 4

Dividing numerator and denominator by

x^{14}

rewrites the integrand as

\frac{\frac{5}{x ^{6}} + \frac{7}{x ^{8}}}{( \frac{1}{x ^{5}} + \frac{1}{x ^{7}} + 2 ) ^{2}}

Putting

t = \frac{1}{x ^{5}} + \frac{1}{x ^{7}} + 2

gives

d t = - (\frac{5}{x ^{6}} + \frac{7}{x ^{8}}) d x

, so

f (x) = - \int \frac{d t}{t ^{2}} = \frac{1}{t} + C = \frac{x ^{7}}{x ^{2} + 1 + 2 x ^{7}} + C

The condition

f (0) = 0

forces

C = 0

, and then

f (1) = \frac{1}{1 + 1 + 2} = \frac{1}{4}

. Comparing with

f (1) = \frac{1}{K}

gives

K = 4

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Numerical answer