Question

_____________________!?​

Answer 1

$\huge\star\underline\mathfrak{Answer:-}⋆$

Answer 2

$\int _0^{\frac{1}{4}\sqrt{3}}\frac{2x\arcsin \left(2x\right)}{\sqrt{1-4x^2}}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \int _0^{\frac{1}{4}\sqrt{3}}\frac{x\arcsin \left(2x\right)}{\sqrt{1-4x^2}}dx$

$\mathrm{Apply\:u-substitution:}\:u=2x$

$=2\cdot \int _0^{\frac{\sqrt{3}}{2}}\frac{u\arcsin \left(u\right)}{4\sqrt{1-u^2}}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \frac{1}{4}\cdot \int _0^{\frac{\sqrt{3}}{2}}\frac{u\arcsin \left(u\right)}{\sqrt{1-u^2}}du$

$\frac{u\arcsin \left(u\right)}{\sqrt{1-u^2}}=\left(u\arcsin \left(u\right)\right)\frac{1}{\sqrt{1-u^2}}$

$=2\cdot \frac{1}{4}\cdot \int _0^{\frac{\sqrt{3}}{2}}u\arcsin \left(u\right)\frac{1}{\sqrt{1-u^2}}du$

$\mathrm{Apply\:Integration\:By\:Parts:}\:u=\arcsin \left(u\right),\:v'=\frac{1}{\sqrt{1-u^2}}u$

$=2\cdot \frac{1}{4}\left[-\arcsin \left(u\right)\sqrt{1-u^2}-\int \:-1du\right]^{\frac{\sqrt{3}}{2}}_0$

$\int \:-1du=-u$

$=2\cdot \frac{1}{4}\left[-\arcsin \left(u\right)\sqrt{1-u^2}-\left(-u\right)\right]^{\frac{\sqrt{3}}{2}}_0$

$\mathrm{Simplify\:}2\cdot \frac{1}{4}\left[-\arcsin \left(u\right)\sqrt{1-u^2}-\left(-u\right)\right]^{\frac{\sqrt{3}}{2}}_0:\quad \frac{1}{2}\left[-\arcsin \left(u\right)\sqrt{1-u^2}+u\right]^{\frac{\sqrt{3}}{2}}_0$

$=\frac{1}{2}\left[-\arcsin \left(u\right)\sqrt{1-u^2}+u\right]^{\frac{\sqrt{3}}{2}}_0$

$\mathrm{Compute\:the\:boundaries}:\quad \left[-\arcsin \left(u\right)\sqrt{1-u^2}+u\right]^{\frac{\sqrt{3}}{2}}_0=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$

$=\frac{1}{2}\left(\frac{\sqrt{3}}{2}-\frac{\pi }{6}\right)$

$\mathrm{Simplify}$

$=\frac{3\sqrt{3}-\pi }{12}$