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

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

Question

(D) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + 1 + \frac{1}{{{x^2}}}}}\,dx} = \int {\frac{{1 - \fra

Vedclass · Accepted Answer

$\frac{1}{2}\log \left( {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + c$

Vedclass · Answer

(D) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + 1 + \frac{1}{{{x^2}}}}}\,dx} = \int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 1}}\,dx}$ Let $t = x + \frac{1}{x}$,then $dt = (1 - \frac{1}{{{x^2}}})\,dx$. Substituting these into the integral: $\int {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2}\log \left| {\frac{{t - 1}}{{t + 1}}} \right| + c$ Substituting $t = x + \frac{1}{x}$ back: $\frac{1}{2}\log \left| {\frac{{x + \frac{1}{x} - 1}}{{x + \frac{1}{x} + 1}}} \right| + c = \frac{1}{2}\log \left( {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + c$.

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

Similar Questions

$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}} = $

There exists $\theta$ such that $a > |\sec \theta|$,then $\int \frac{dx}{1+a \cos x} = $

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

The value of $\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$ is equal to (where $C$ is an arbitrary constant.)

If $\int \operatorname{cosec}^5 x \, dx = \alpha \cot x \operatorname{cosec} x \left(\operatorname{cosec}^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$,where $\alpha, \beta \in R$ and $C$ is the constant of integration,then the value of $8(\alpha + \beta)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module