If f(x) = int frac{5x^8 + 7x^6}{(x^2 + 1 + 2x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} dx$.Divide the numerator and denominator by $x^{14}$ inside the integral:$f(x) = \int \f

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} dx$.Divide the numerator and denominator by $x^{14}$ inside the integral:$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^2} dx$.Let $t = x^{-5} + x^{-7} + 2$. Then $dt = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = dt$.Substituting this into the integral:$f(x) = -\int \frac{dt}{t^2} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$.Simplifying the expression:$f(x) = \frac{x^7}{1 + x^2 + 2x^7} + C$.Given $f(0) = 0$,we have $0 = \frac{0}{1} + C$,so $C = 0$.Thus,$f(x) = \frac{x^7}{x^2 + 1 + 2x^7}$.Evaluating at $x = 1$:$f(1) = \frac{1^7}{1^2 + 1 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

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

Similar Questions

$\int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$ is equal to

$\int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx=$

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$

$\int {\frac{{{e^{m{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $ is equal to

If $\int \sin ^5 x \, dx = \frac{-\cos ^5 x}{5} + a \cos ^3 x + b \cos x + c$,then $a + b =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module