If f(x)=int frac{x^2 , dx}{(1+x^2)(1+sqrt{1+x | vedclass

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

Question

(B) Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.Then,$f(x) = \int \frac{	an^2 	heta \sec^2 	heta \, d	heta}{\sec^2 	heta(1+

Vedclass · Accepted Answer

$\log (1+\sqrt{2})-\frac{\pi}{4}$

Vedclass · Answer

(B) Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.Then,$f(x) = \int \frac{	an^2 	heta \sec^2 	heta \, d	heta}{\sec^2 	heta(1+\sec 	heta)} = \int \frac{	an^2 	heta \, d	heta}{1+\sec 	heta}$.Since $	an^2 	heta = \sec^2 	heta - 1$,we have $\int \frac{\sec^2 	heta - 1}{1+\sec 	heta} \, d	heta = \int (\sec 	heta - 1) \, d	heta$.Integrating gives $f(x) = \log |\sec 	heta + 	an 	heta| - 	heta + c$.Substituting back $x = 	an 	heta$ and $\sec 	heta = \sqrt{1+x^2}$,we get $f(x) = \log |x + \sqrt{1+x^2}| - 	an^{-1} x + c$.Given $f(0) = 0$,we have $\log |0 + 1| - 	an^{-1}(0) + c = 0$,which implies $c = 0$.Thus,$f(x) = \log |x + \sqrt{1+x^2}| - 	an^{-1} x$.Evaluating at $x=1$,$f(1) = \log |1 + \sqrt{2}| - 	an^{-1}(1) = \log (1+\sqrt{2}) - \frac{\pi}{4}$.

If $f(x)=\int \frac{x^2 \, dx}{(1+x^2)(1+\sqrt{1+x^2})}$ and $f(0)=0$,then $f(1)$ is

Similar Questions

$\int {\frac{{\sec x(1 + \tan x)dx}}{{({e^{ - x}} + \sec x)}}} = f(x) + C$ where $f(0) = \ln 2$,then $f\left( {\frac{\pi }{4}} \right)$ is -

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

The value of $\int \frac{\sin x - \cos x}{\sin x + \cos x} \,dx$ is

$\int \frac{dx}{\cos^2(x) + \sin(2x)} = $

$\int \frac{e^x(1+x)}{\sin ^2(x \cdot e^x)} dx = $ . . . . . . $+ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module