If int frac{2 x^{12}+5 x^9}{left(x^5+x^3+1rig | vedclass

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

Question

(B) Given the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} d x$. Divide the numerator and denominator by $x^{15}$ inside the int

Vedclass · Accepted Answer

$\frac{1}{9}$

Vedclass · Answer

(B) Given the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} d x$. Divide the numerator and denominator by $x^{15}$ inside the integral: $I = \int \frac{2 x^{-3} + 5 x^{-6}}{(1 + x^{-2} + x^{-5})^3} d x$. Let $t = 1 + x^{-2} + x^{-5}$. Then $dt = (-2x^{-3} - 5x^{-6}) dx$, which implies $-(2x^{-3} + 5x^{-6}) dx = dt$. Substituting this into the integral: $I = -\int \frac{dt}{t^3} = -\int t^{-3} dt = -\frac{t^{-2}}{-2} + C = \frac{1}{2t^2} + C$. Comparing this with $\frac{1}{2} f(x) + C$, we get $f(x) = \frac{1}{t^2} = \frac{1}{(1 + x^{-2} + x^{-5})^2} = \frac{x^{10}}{(x^5 + x^3 + 1)^2}$. Now, calculate $f(1) - f(0)$. $f(1) = \frac{1^10}{(1^5 + 1^3 + 1)^2} = \frac{1}{3^2} = \frac{1}{9}$. For $f(0)$, the expression $\frac{x^{10}}{(x^5 + x^3 + 1)^2}$ at $x=0$ is $0$. Thus, $f(1) - f(0) = \frac{1}{9} - 0 = \frac{1}{9}$.

If $\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} d x=\frac{1}{2} f(x)+C$, then $f(1)-f(0)=$

Similar Questions

The value of $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is

If $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$, then $f(x)$ is equal to

$\int \frac{1}{\sin (x-a) \sin x} \,d x=$

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

If $\int {\frac{{\cos 4x + 1}}{{\cot x - \tan x}}} dx = k\,\cos 4x + c$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module