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1)  why there is a square root sign and then a square?  it is not clear.

\int\limits^{}_{} {[\sqrt{x^2}-\frac{1}{\sqrt{x^2}}]^2} \, dx \\\\=\int\limits^{}_{} {(x-\frac{1}{x})^2} \, dx \\\\=\int\limits^{}_{} {(x^2+\frac{1}{x^2}-2)} \, dx \\\\x^3/3-\frac{1}{x}-2x+K

2)  Is that x^2 or  x^4.

f(x)=(x^4+5x^2-4)/x^2=x^2+5-4x^{-2}\\\\\int\limits^{}_{} {f(x)} \, dx =x^3/3+5x+4/x

f(x)=(1-x)\sqrt{x}=x^{\frac{1}{2}}-x^{\frac{3}{2}}\\\\\int\limits^{}_{} {f(x)} \, dx =\frac{1}{(\frac{3}{2})}x^{(\frac{3}{2})}-\frac{1}{(\frac{5}{2})}x^{(\frac{3}{2}+1)}


There is no simple answer to this... x^x = e^{x ln x}
Expand this in power series. like e^x.
So f9x) =  1 + x ln x + (x  ln x)^2 /2  +  (x ln x)^3 / 6 + ....
Try integrating each term separately.
Even those integrals are not easy to obtain..  It is simpler to evaluate  definite integrals with x ranging fro m  0 to 1.. and indefinite integral is not easy to find.

put cos x = y.    so - sinx dx = dy.   then,
 integral of  - Cos y  dy =  - Sin y  = - Sin (cosx)  + K


2 x³/3 + 3 cos x + 5 x(³/²)  + K

2 x² / 2   - 3 sin x + e^x + K

1 5 1
is that e^(2x+3) or x^(2x+3) ?
Thnx... Sir i got confused in the fourth 1 , :)
the first question too.. u write differently.. i did not get it.. please write answer again for them, i suppose u have the answers saved with you somewhere..
sir i didnt answer the 1st one , i asked about it
that ,