不定積分∫dx/(x^4+7)的計算步驟
主要內容:
本題通過湊分、換元、裂項、反正切函數導數、冪函數導數等方法和知識,介紹不定積分∫dx/(x^4+7)的主要計算步驟。
※.主要步驟
∫dx/(x^4+7)
=∫dx/[7 (x^4/7+1)
=(1/7)∫dx/[(√x/√7)^4+1]
設√x/√7=t,則x=√7t,代入得:
=(1/7)∫d(√7t )/(t^4+1),
=(1/√7)∫dt/(t^4+1)
=(1/2√7)∫[(t^2+1)-(t^2-1)]dt/(t^4+1),此步驟爲對分子進行等量變換,
=(1/2√7)∫(t^2+1)dt/(t^4+1)- (1/2√7)∫(t^2-1)dt/(t^4+1),此步驟爲裂項,
=(1/2√7)∫(t^2+1)dt/(t^4+1)- (1/2√7)∫(t^2-1)dt/(t^4+1),兩項分子分母同時除以t^2得,
=(1/2√7)∫[1+(1/t^2)]dt/[t^2+(1/t^2)]- (1/2√7)∫[1-(1/t^2)]dt/[t^2+(1/t^2)],
=(1/2√7)∫d(t-1/t)/[t^2+(1/t^2)]- (1/2√7)∫d(t+1/t)/[t^2+(1/t^2)],
此步驟爲分子湊分法,
=(1/2√7)∫d(t-1/t)/[(t-1/t)^2+2]-(1/2√7)∫d(t+1/t)/[(t+1/t)^2-2],此步驟爲根據分子對分母進行配方計算,
=(1/2√7)∫d(t-1/t)/2[(t-1/t)^2/2+1]-(1/2√7)∫d(t+1/t)/{[(t+1/t)-√2][ (t+1/t)+√2]},
此步驟前者對分母提取公因式2,後者使用平方差公式,即:
=(1/2√7)arctan[(t-1/t)/√2]- (1/4√14){∫d(t+1/t)/[(t+1/t)-√2]-∫d(t+1/t)/[(t+1/t)+√2]},
=(1/2√7)arctan[(t-1/t)/√2]- (1/4√14)ln|[(t+1/t)-√2]/ [(t+1/t)+√2]|+C.
由t=√x/√7知1/t=√7/√x,則:
t-1/t=(x^2-7)/√7x,t+1/t=(x^2+7)/√7x代入上式,有,
=(1/2√7)arctan[(x^2-7)/√2x]-(1/4√14)ln|[(x^2+7)/√7x-√2]/ [(x^2+7)/√7x+√2]|+C,
對後者進行等量變形,則:
所求式
=(1/2√7)arctan[(x^2-7)/√14x]-(1/4√14)ln|[(x^2+7)-√14x]/ [(x^2+7)+√14x]|+C.