Question

1. ∫[√$x^{2}$-(1\√$x^{2}$)]² dx
2.∫ $( x^{2} +5 x^{2} -4)/ x^{2}$ dx
3. ∫$(1-x) \sqrt{x} dx$
4. ∫[tex] x^{2x+3}
[/tex]
5. ∫sinx.sin(cosx)dx
6. ∫2x²-3sinx+5$\sqrt{x}$
7. ∫2x-3cosx+e^x[/tex]

Answer 1

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.

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

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

4)

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.


5)
put cos x = y.    so - sinx dx = dy.   then,
 integral of  - Cos y  dy =  - Sin y  = - Sin (cosx)  + K

6)

2 x³/3 + 3 cos x + 5 x(³/²)  + K

7)
2 x² / 2   - 3 sin x + e^x + K