For real numbers alpha, beta, gamma and delta | vedclass

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

Question

(A) Let $I = \int \frac{(x^2-1) dx}{(x^4+3x^2+1) 	an^{-1}(x+1/x)} + \int \frac{dx}{x^4+3x^2+1}$.Divide numerator and denominator by $x^2$ in both int

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let $I = \int \frac{(x^2-1) dx}{(x^4+3x^2+1) 	an^{-1}(x+1/x)} + \int \frac{dx}{x^4+3x^2+1}$.Divide numerator and denominator by $x^2$ in both integrals.$I = \int \frac{(1-1/x^2) dx}{((x+1/x)^2+1) 	an^{-1}(x+1/x)} + \frac{1}{2} \int \frac{(1+1/x^2) - (1-1/x^2) dx}{(x^2+1/x^2+3)}$.Let $t = 	an^{-1}(x+1/x)$,then $dt = \frac{1}{1+(x+1/x)^2} (1-1/x^2) dx$.$I = \int \frac{dt}{t} + \frac{1}{2} \int \frac{1+1/x^2}{(x-1/x)^2+5} dx - \frac{1}{2} \int \frac{1-1/x^2}{(x+1/x)^2+1} dx$.$I = \log_e |t| + \frac{1}{2} \int \frac{dy}{y^2+5} - \frac{1}{2} \int \frac{dz}{z^2+1}$,where $y=x-1/x$ and $z=x+1/x$.$I = \log_e |	an^{-1}(x+1/x)| + \frac{1}{2\sqrt{5}} 	an^{-1}(\frac{x-1/x}{\sqrt{5}}) - \frac{1}{2} 	an^{-1}(x+1/x) + C$.Comparing with the given form,$\alpha=1, \beta=\frac{1}{2\sqrt{5}}, \gamma=\frac{1}{\sqrt{5}}, \delta=-\frac{1}{2}$.Then $10(\alpha+\beta\gamma+\delta) = 10(1 + \frac{1}{2\sqrt{5}} \cdot \frac{1}{\sqrt{5}} - \frac{1}{2}) = 10(1 + \frac{1}{10} - \frac{1}{2}) = 10(\frac{10+1-5}{10}) = 6$.

Similar Questions

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$ then $\frac{m-l}{r}=$

$\int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta$ equals:

$\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx} = $

If $\frac{3 \pi}{2} < x < \frac{5 \pi}{2}$ and $\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) \, dx = f(x) + c$ where $c$ is the constant of integration, then $f\left(\frac{\pi}{3}\right) - f(0) =$

$\begin{aligned} & \text{If } 5(f(x))^2 = x f(x) + 30 \text{ and } \\ & \int \frac{3 x^3 + (1 - 30 x^2) f(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx \\ & = \frac{A}{B x^3 + D f(x)} + C, \text{ then } A + B + D = \end{aligned}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module