If int frac{(x-1) dx}{(x+1) sqrt{x^3+x^2+x}} | vedclass

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

Question

(C) Let $I = \int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} dx$. Divide the numerator and denominator by $x$ inside the square root and simplify: $I = \int \

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(C) Let $I = \int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} dx$. Divide the numerator and denominator by $x$ inside the square root and simplify: $I = \int \frac{x-1}{(x+1) \sqrt{x^2(x+1+1/x)}} dx = \int \frac{x-1}{x(x+1) \sqrt{x+1+1/x}} dx$. Multiply numerator and denominator by $x$: $I = \int \frac{1-1/x^2}{(x+1/x+2) \sqrt{x+1/x+1}} dx$. Let $t = \sqrt{x+1/x+1}$, then $t^2 = x+1/x+1$. Differentiating both sides, $2t dt = (1-1/x^2) dx$. Substituting these into the integral: $I = \int \frac{2t dt}{(t^2+1)t} = 2 \int \frac{dt}{t^2+1} = 2 	an^{-1}(t) + C$. Thus, $f(x) = 2 	an^{-1}(\sqrt{x+1/x+1})$. Evaluating at $x=1$: $f(1) = 2 	an^{-1}(\sqrt{1+1+1}) = 2 	an^{-1}(\sqrt{3}) = 2 	imes \frac{\pi}{3} = \frac{2 \pi}{3}$.

If $\int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}} = f(x) + C$, then $f(1) =$

Similar Questions

$\int \frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)} \times [\log g(x)-\log f(x)] \, dx$ is equal to

$\int \frac{\sec^2 x}{\log ((\tan x)^{\tan x})} dx = $

$\int \frac{d x}{(x+2) \sqrt{x+1}} = $

The value of $\int \frac{dx}{x^2(x^4+1)^{3/4}}$ is

The integral $\int \sec^{2/3} x \csc^{4/3} x \, dx$ is equal to: (Here $C$ is a constant of integration)

Vedclass Test Series

Exam Paper Generator

Online Exam Module