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

$\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x=$

$\int \frac{1}{x \cos^2(1 + \log x)} \, dx = $

If $f(x) = \sqrt{\tan x}$ and $g(x) = \sin x \cdot \cos x$,then $\int \frac{f(x)}{g(x)} dx$ is equal to (where $C$ is a constant of integration).

If $\int \frac{x+1}{\sqrt{2x-1}} \, dx = f(x) \sqrt{2x-1} + C$,where $C$ is an arbitrary constant,then $f(x)$ is equal to

If $\int {f(x)\,dx = f(x)} ,$ then $\int {{{\left[ {f(x)} \right]}^2}\,dx}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module