int {frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1 | vedclass

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

Question

(A) To solve the integral $\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx}$,we first simplify the integrand by factoring the numerator.We know

Vedclass · Accepted Answer

$\frac{1}{3}{x^3} + \frac{1}{2}{x^2} + x + c$

Vedclass · Answer

(A) To solve the integral $\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx}$,we first simplify the integrand by factoring the numerator.We know that ${x^4} + {x^2} + 1 = ({x^2} + 1)^2 - {x^2} = ({x^2} + 1 - x)({x^2} + 1 + x)$.Substituting this into the integral,we get:$\int {\frac{({x^2} - x + 1)({x^2} + x + 1)}{{{x^2} - x + 1}}\,dx} = \int {({x^2} + x + 1)\,dx}$.Now,integrate term by term:$\int {x^2\,dx} + \int {x\,dx} + \int {1\,dx} = \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + c$.Thus,the correct option is $A$.

$\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx} = $

Similar Questions

$\int \frac{x^5+1}{x+1} \, dx = $ . . . . . . $+ c$.

$\int \frac{1}{16 x^{2}+9} d x$ is equal to

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$ is equal to

$\int \tan^2 x \, dx$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module