Question

$\footnotesize \rm\displaystyle \rm{\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}}$​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\sqrt[4]{1-8{x}^{2}+8{x}^{4}-4x\sqrt{{x}^{2}-1}+8{x}^{3}\sqrt{{x}^{2}-1}}\,dx$

$\displaystyle=\int\sqrt[4]{4{x}^{4}-4{x}^{2}+1+4{x}^{4}-4{x}^{2}+4x\sqrt{{x}^{2}-1}\left(2{x}^{2}-1\right)}\,dx$

$\displaystyle=\int\sqrt[4]{\left(2{x}^{2}-1\right)^{2}+4{x}^{2}\left({x}^{2}-1\right)+4x\sqrt{{x}^{2}-1}\left(2{x}^{2}-1\right)}\,dx$

$\displaystyle=\int\sqrt[4]{\left(2{x}^{2}-1\right)^{2}+2\cdot2x\sqrt{{x}^{2}-1}\cdot\left(2{x}^{2}-1\right)+\left(2x\sqrt{{x}^{2}-1}\right)^{2}}\,dx$

The stuff inside the 4th root is in the form of (a²+2ab+b²)=(a+b)²

So,

$\displaystyle=\int\sqrt[4]{\left(2{x}^{2}-1+2x\sqrt{{x}^{2}-1}\right)^{2}}\,dx$

$\displaystyle=\int\sqrt[4]{\left({x}^{2}+{x}^{2}-1+2x\sqrt{{x}^{2}-1}\right)^{2}}\,dx$

$\displaystyle=\int\sqrt[4]{\left(\left(x\right)^{2}+\left(\sqrt{{x}^{2}-1}\right)^{2}+2\cdot\,x\cdot\sqrt{{x}^{2}-1}\right)^{2}}\,dx$

Again, the same form as before,

$\displaystyle=\int\sqrt[4]{\left(\left(x+\sqrt{{x}^{2}-1}\right)^{2}\right)^{2}}\,dx$

$\displaystyle=\int\sqrt[4]{\left(x+\sqrt{{x}^{2}-1}\right)^{4}}\,dx$

$\displaystyle=\int\left(x+\sqrt{{x}^{2}-1}\right)\,dx$

$\displaystyle=\int\,x\,dx+\int\sqrt{{x}^{2}-1}\,dx$

$\displaystyle=\dfrac{{x}^{2}}{2}+\dfrac{x}{2}\sqrt{{x}^{2}-1}-\dfrac{1}{2}\ln\left|x+\sqrt{{x}^{2}-1}\right|+C$