The value of the integral int_{-1}^1 left( fr | vedclass

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

Question

(B) Let $I = \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$. Since the denominator $x^2 + 2|x| + 1 = (|x| + 1)^2$ is an even function, we can sp

Vedclass · Accepted Answer

$2 \log_e 2$

Vedclass · Answer

(B) Let $I = \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$. Since the denominator $x^2 + 2|x| + 1 = (|x| + 1)^2$ is an even function, we can split the integral into two parts: $I = \int_{-1}^0 \frac{x^3 - x + 1}{(|x| + 1)^2} dx + \int_0^1 \frac{x^3 + x + 1}{(|x| + 1)^2} dx$. For the first part, let $x = -t$, then $dx = -dt$. As $x$ goes from $-1$ to $0$, $t$ goes from $1$ to $0$. $\int_1^0 \frac{-t^3 + t + 1}{(t + 1)^2} (-dt) = \int_0^1 \frac{-t^3 + t + 1}{(t + 1)^2} dt$. Thus, $I = \int_0^1 \frac{x^3 + x + 1 - x^3 + x + 1}{(x + 1)^2} dx = \int_0^1 \frac{2x + 2}{(x + 1)^2} dx = \int_0^1 \frac{2(x + 1)}{(x + 1)^2} dx = \int_0^1 \frac{2}{x + 1} dx$. $I = 2 [\log_e(x + 1)]_0^1 = 2(\log_e 2 - \log_e 1) = 2 \log_e 2$.

The value of the integral $\int_{-1}^1 \left( \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \right) dx$ is equal to:

Similar Questions

$\int_1^2 \frac{x^4-1}{x^6-1} d x=$

Dividing the interval $[0, 6]$ into $6$ equal parts and by using the trapezoidal rule,the value of $\int_0^6 x^3 \, dx$ is approximately:

$\int_0^{2\pi} (\sin x + \cos x) \, dx = $

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

$\int_0^1 {{\cos }^{ - 1}}x\,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module