A question of indefinite integrals---(1)

As follows:

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2007-10-31 01:26:24 補充：
If we solve sin[tan^-1 (x/2)] by means of Pythagorus theorem, we have it equal to x/√(x^2 4)Therefore an alternative expression for the solution is:x/[4√(x^2 4)] C

Both answers are correct!

given x = 2 tan (theta)

∫ [ 1 / ( x ² + 4 ) ^ ( 3 / 2 ) ] ( d x )

= 1/4 * sin (theta) + C

= 1/4 * (1 + tan²(theta) ) ^ (-1/2)

= 1/4 * (1 + x² / 4) ^ (-1/2)

= 1/4 * 2 * (4 + x²) (-1/2)

= (1/2) ( x ² + 4 ) ^ (- 1 / 2 ) + C

為方便睇..我將 1 / x = x ^ (-1)....係相等的
1 / ( x ² + 4 ) ^ ( 3 / 2 ) = ( x ² + 4 ) ^ (- 3 / 2 )

∫ [( x ² + 4 ) ^ (- 3 / 2 )] (dx)

=∫ [( x ² + 4 ) ^ (- 3 / 2 )] [( 1 / 2) d ( x ² + 4 ) ]

= (1 / 2)∫ [( x ² + 4 ) ^ (- 3 / 2 )] d ( x ² + 4 )

= {(1 / 2) [( x ² + 4 ) ^ (- 1 / 2 )]} (- 1 / 2)................

= - ( x ² + 4 ) ^ (- 1 / 2 )...

2007-10-31 00:28:50 補充：
加 constantans : - ( x ² + 4 ) ^ (- 1 / 2 ) + c