int_1^2 frac{x^4-1}{x^6-1} d x= | vedclass

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

Question

(A) Let $I = \int_1^2 \frac{x^4-1}{x^6-1} d x$. We simplify the integrand: $\frac{x^4-1}{x^6-1} = \frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} = \frac{x^

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} 	an ^{-1}\left(\frac{\sqrt{3}}{2}ight)$

Vedclass · Answer

(A) Let $I = \int_1^2 \frac{x^4-1}{x^6-1} d x$. We simplify the integrand: $\frac{x^4-1}{x^6-1} = \frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} = \frac{x^2+1}{x^4+x^2+1}$. Note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. Using partial fractions: $\frac{x^2+1}{(x^2+x+1)(x^2-x+1)} = \frac{1}{2} \left( \frac{1}{x^2+x+1} + \frac{1}{x^2-x+1} ight)$. Integrating: $I = \frac{1}{2} \int_1^2 \left( \frac{1}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} ight) d x$. $I = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \left[ 	an^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}ight) + 	an^{-1}\left(\frac{x-1/2}{\sqrt{3}/2}ight) ight]_1^2$. $I = \frac{1}{\sqrt{3}} \left[ 	an^{-1}\left(\frac{2x+1}{\sqrt{3}}ight) + 	an^{-1}\left(\frac{2x-1}{\sqrt{3}}ight) ight]_1^2$. Evaluating at limits: $I = \frac{1}{\sqrt{3}} \left[ (	an^{-1}(\frac{5}{\sqrt{3}}) + 	an^{-1}(\sqrt{3})) - (	an^{-1}(\sqrt{3}) + 	an^{-1}(\frac{1}{\sqrt{3}})) ight]$. $I = \frac{1}{\sqrt{3}} \left[ 	an^{-1}(\frac{5}{\sqrt{3}}) - 	an^{-1}(\frac{1}{\sqrt{3}}) ight]$. Using $	an^{-1} A - 	an^{-1} B = 	an^{-1}(\frac{A-B}{1+AB})$: $I = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{5/\sqrt{3} - 1/\sqrt{3}}{1 + 5/3} ight) = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{4/\sqrt{3}}{8/3} ight) = \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{\sqrt{3}}{2} ight)$.

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

Similar Questions

$\int_{0}^{5} \frac{d x}{x^{2}+2 x+10} = $

$\int_2^3 \frac{\log x}{x} d x=$

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x) dx$ is equal to:

Let $f$ be a real-valued continuous function defined on the positive real axis such that $g(x) = \int_0^x t f(t) dt$. If $g(x^3) = x^6 + x^7$,then the value of $\sum_{r=1}^{15} f(r^3)$ is:

Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module