intlimits_{ - 1}^1 {frac{{{x^3} + |x| + 1}}{{ | vedclass

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

Question

(B) Let $I = \int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx$. Since the denominator ${x^2} + 2|x| + 1 = (|x| + 1)^2$ is an eve

Vedclass · Accepted Answer

$a = 2; b = 0$

Vedclass · Answer

(B) Let $I = \int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx$. Since the denominator ${x^2} + 2|x| + 1 = (|x| + 1)^2$ is an even function,we can split the integral into odd and even parts. $I = \int\limits_{ - 1}^1 {\frac{{{x^3}}}{{(|x| + 1)^2}}} dx + \int\limits_{ - 1}^1 {\frac{{|x| + 1}}{{(|x| + 1)^2}}} dx$. The first part $\int\limits_{ - 1}^1 {\frac{{{x^3}}}{{(|x| + 1)^2}}} dx$ is an integral of an odd function over a symmetric interval $[-1, 1]$,so it equals $0$. Thus,$I = \int\limits_{ - 1}^1 {\frac{1}{{|x| + 1}}} dx$. Since $\frac{1}{{|x| + 1}}$ is an even function,$I = 2 \int\limits_0^1 {\frac{1}{{x + 1}}} dx$. $I = 2 [\ln(x + 1)]_0^1 = 2(\ln 2 - \ln 1) = 2 \ln 2$. Comparing $2 \ln 2$ with $a \ln 2 + b$,we get $a = 2$ and $b = 0$.

$\int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx = a \ln 2 + b$,then:

Similar Questions

If $[x]$ is the greatest integer $\leq x$,then $\pi^{2} \int_{0}^{2}\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x$ is equal to :

$\int_0^1 \frac{8 \log (1+x)}{1+x^2} dx =$

If $[ \cdot ]$ represents the greatest integer function,then $\int_{-1}^1 (x[1+\sin(\pi x)]+1) dx = $

$\int_{-2}^{2} |x| \, dx = $

If $\int_0^\pi \frac{d x}{1+2 \sin ^2 x}=k$,then the greatest integer less than or equal to $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module