int_{-1}^1 frac{x^3+|x|+1}{x^2+2|x|+1} dx is | vedclass

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

Question

(B) Let $I = \int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$. We can split the integral into two parts: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int

Vedclass · Accepted Answer

$2 \log 2$

Vedclass · Answer

(B) Let $I = \int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$. We can split the integral into two parts: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} dx$. Let $f(x) = \frac{x^3}{x^2+2|x|+1}$. Since $f(-x) = \frac{(-x)^3}{(-x)^2+2|-x|+1} = -\frac{x^3}{x^2+2|x|+1} = -f(x)$,the function is odd. Therefore,$\int_{-1}^1 f(x) dx = 0$. Now,consider the second part $I_2 = \int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} dx$. Since the integrand is an even function,$I_2 = 2 \int_0^1 \frac{x+1}{x^2+2x+1} dx$. Simplifying the integrand: $I_2 = 2 \int_0^1 \frac{x+1}{(x+1)^2} dx = 2 \int_0^1 \frac{1}{x+1} dx$. Evaluating the integral: $I_2 = 2 [\ln |x+1|]_0^1 = 2 (\ln 2 - \ln 1) = 2 \ln 2$. Thus,$I = 0 + 2 \ln 2 = 2 \ln 2$.

$\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$ is equal to

Similar Questions

Suppose $f$ is such that $f(-x) = -f(x)$ for every real $x$ and $\int_{0}^{1} f(x) dx = 5$,then $\int_{-1}^{0} f(t) dt = $

Evaluate the integral: $\int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$

Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

The value of the integral $\int_{4}^{10} \frac{[x^2]}{[(x-14)^2] + [x^2]} dx$,where $[x]$ denotes the greatest integer function,is

The value of $\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module