If int frac{(x^2-1)}{(x+1)^2 sqrt{x(x^2+x+1)} | vedclass

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

Question

(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$. Consider the derivative of the right-hand side: $\frac{d}{dx} \left( A 	an^{-1} \sqrt{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$. Consider the derivative of the right-hand side: $\frac{d}{dx} \left( A 	an^{-1} \sqrt{\frac{x^2+x+1}{x}} ight) = A \cdot \frac{1}{1 + \frac{x^2+x+1}{x}} \cdot \frac{d}{dx} \left( \sqrt{\frac{x^2+x+1}{x}} ight)$. $= A \cdot \frac{x}{x^2+2x+1} \cdot \frac{1}{2 \sqrt{\frac{x^2+x+1}{x}}} \cdot \left( \frac{x(2x+1) - (x^2+x+1)}{x^2} ight)$. $= A \cdot \frac{x}{(x+1)^2} \cdot \frac{\sqrt{x}}{2 \sqrt{x^2+x+1}} \cdot \left( \frac{2x^2+x-x^2-x-1}{x^2} ight)$. $= A \cdot \frac{x}{(x+1)^2} \cdot \frac{\sqrt{x}}{2 \sqrt{x^2+x+1}} \cdot \frac{x^2-1}{x^2}$. $= A \cdot \frac{x^2-1}{2(x+1)^2 \sqrt{x(x^2+x+1)}}$. Comparing this with the integrand, we have $\frac{A}{2} = 1$, which implies $A = 2$.

If $\int \frac{(x^2-1)}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx = A \tan^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}\right) + C$, where $C$ is a constant, then $A$ equals to

Similar Questions

If $\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} $ where $C$ is a constant of integration,then the function $f(x)$ is equal to

$\int \sin ^{-1}\left(\sqrt{\frac{x-a}{x}}\right) d x=$

Integrate the function: $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$

Let $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. If $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$,where $b, c \in \mathbb{N}$,then $3(b+c)$ is equal to

Integrate the function: $\frac{x+3}{x^{2}-2x-5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module