int {frac{{xdx}}{{sqrt {1 + {x^2} + sqrt {{{l | vedclass

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

Question

(D) Let $I = \int \frac{xdx}{\sqrt{1 + x^2 + (1 + x^2)^{3/2}}}$.Substitute $1 + x^2 = t^2$,then $2xdx = 2tdt$,which implies $xdx = tdt$.Substituting t

Vedclass · Accepted Answer

$2\left( {\sqrt {1 + \sqrt {1 + {x^2}} } } \right) + C$

Vedclass · Answer

(D) Let $I = \int \frac{xdx}{\sqrt{1 + x^2 + (1 + x^2)^{3/2}}}$.Substitute $1 + x^2 = t^2$,then $2xdx = 2tdt$,which implies $xdx = tdt$.Substituting these into the integral:$I = \int \frac{tdt}{\sqrt{t^2 + (t^2)^{3/2}}} = \int \frac{tdt}{\sqrt{t^2 + t^3}}$.Since $t > 0$,$\sqrt{t^2 + t^3} = t\sqrt{1 + t}$.$I = \int \frac{tdt}{t\sqrt{1 + t}} = \int \frac{dt}{\sqrt{1 + t}}$.Integrating with respect to $t$:$I = 2\sqrt{1 + t} + C$.Substituting back $t = \sqrt{1 + x^2}$:$I = 2\sqrt{1 + \sqrt{1 + x^2}} + C$.

$\int {\frac{{xdx}}{{\sqrt {1 + {x^2} + \sqrt {{{\left( {1 + {x^2}} \right)}^3}} } }}} $ is equal to (where $C$ denotes constant of integration)

Similar Questions

The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} d x$ is equal to (Where $C$ is a constant of integration).

Integrate the function $\frac{(\log x)^{2}}{x}$.

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

$\int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx =$

The value of $\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$ for $x>0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module