If int frac{sqrt{x}}{x(x+1)} d x = k tan^{-1} | vedclass

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

Question

(D) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$. Substitute $x = 	an^2 	heta$,which implies $dx = 2 	an 	heta \sec^2 	heta d	heta$. Then,$\sqrt{x}

Vedclass · Accepted Answer

$(2, \sqrt{x})$

Vedclass · Answer

(D) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$. Substitute $x = 	an^2 	heta$,which implies $dx = 2 	an 	heta \sec^2 	heta d	heta$. Then,$\sqrt{x} = 	an 	heta$. Substituting these into the integral: $I = \int \frac{	an 	heta}{	an^2 	heta (	an^2 	heta + 1)} \cdot (2 	an 	heta \sec^2 	heta) d	heta$. Since $	an^2 	heta + 1 = \sec^2 	heta$,the expression simplifies to: $I = \int \frac{	an 	heta}{	an^2 	heta \sec^2 	heta} \cdot (2 	an 	heta \sec^2 	heta) d	heta$. $I = \int \frac{2 	an^2 	heta \sec^2 	heta}{	an^2 	heta \sec^2 	heta} d	heta = \int 2 d	heta$. $I = 2	heta + C$. Since $x = 	an^2 	heta$,we have $	heta = 	an^{-1}(\sqrt{x})$. Therefore,$I = 2 	an^{-1}(\sqrt{x}) + C$. Comparing this with $k 	an^{-1} m$,we get $k = 2$ and $m = \sqrt{x}$. Thus,$(k, m) = (2, \sqrt{x})$.

If $\int \frac{\sqrt{x}}{x(x+1)} d x = k \tan^{-1} m$,then $(k, m)$ is

Similar Questions

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$ for $n \geq 2$ and $g(x) = \underbrace{(f \circ f \circ \ldots \circ f)}_{n \text{ times }}(x)$. Then $\int x^{n-2} g(x) \, dx$ equals

$\int \frac{\sin x}{3+4 \cos ^2 x} d x$

$\int \cos ^3 x \cdot e^{\log (\sin x)} d x=$

If $I = \int \frac{\sin^2 x - 1}{2x \sin^2 x + \sin 2x} dx$,then. . . . . $( \sin x \neq 0)$ (where $c$ is a constant of integration)

$\int \frac{dx}{5 + 4\cos x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module