指對數微積分(exp x 二次方)(有點難10點!!)

Question

1.f(x)=exp x^2 ，d^3f(x)/dx^3 =??

2.∫secxdx 範圍0~π/4 =?

3.∫(2x+1/2x^2+1) dx =?

4.∫x exp x^2 dx 範圍1~2^1/2 =?

5.∫x 2^x^2 dx 範圍也是1~2^1/2=?

恭請高手幫幫忙...>_<

我第三題打錯了= ="想說怎麼算的那麼難
∫(2x+1/2x^2+x+1) dx =
中間少加一個x

等等等...這個才是正確(要問的)
.∫(2x+1/2x^2+x-1) dx =?

Answer 1

1. f(x) = exp(x2)d3f(x)/dx3= (exp(x2))'''= (2xexp(x2))''= (2x(2xexp(x2)+2exp(x2))'= (4x2exp(x2)+2exp(x2))' = ((4x2+2)exp(x2))'= (4x2+2)(2x)exp(x2)+8xexp(x2)= (8x3+12x)exp(x2)2. ∫0π/4 secx dx= ln∣secx + tanx∣] 0π/4 = ln ∣sec(π/4) + tan(π/4)∣- ln∣sec 0 + tan 0∣= ln(√2 + 1)3. 令x= (1/√2)tanθ , -π/2<θ<π/2則dx=(1/√2)sec2θdθ2x2+1 = 2(1/2)tan2θ+1= tan2θ+1 = sec2θ2x+1 = √2tanθ+1∫(2x+1)/(2x2+1) dx=∫((√2tanθ+1)/sec2θ)(1/√2)sec2θdθ=∫tanθ+(1/√2) dθ= - ln∣cosθ∣+(1/√2)θ+c= - ln(1/√(2x2+1))+(1/√2)tan-1(√2 x)+c= ln(√(2x2+1))+(1/√2)tan-1(√2 x)+c4. ∫1√2 xexp(x2)dx= ½∫1√2 exp(x2)(2x)dx= ½ exp(x2)∣1√2 = ½ (e2 - e)5. ∫1√2 x2x^2dx=∫1√2 2x^2-1(2x)dx= 2x^2-1/ln2∣1√2= 2/ln2 - 1/ln2= 1/ln2

2006-06-29 09:54:36 補充：
上面再補充會超過2000,只能補在這了
∫(2x+1)/(2x²+x-1) dx
=∫(2x+1)/((2x-1)(x+1)) dx
=∫(4/3)/(2x-1) + (1/3)/(x+1) dx
=(2/3)∫2/(2x-1) dx +(1/3)∫1/(x+1) dx
=(2/3) ln∣2x-1∣+ (1/3)ln∣x+1∣+c