Question

∫(2x+3)/(x^2+3x+1)dx和∫1/(x^2+3x+1)dx结果一样吗?

Answer 1

∫(2x+3)/(x²+3x+1)dx和∫1/(x²+3x+1)dx结果一样吗?

解：∫(2x+3)/(x²+3x+1)dx=∫d(x²+3x+1)/(x²+3x+1)=ln∣x²+3x+1∣+C...............(1)
∫1/(x²+3x+1)dx=∫dx/[(x+3/2)²-5/4]=(4/5)∫dx/{[2(x+3/2)/√5]²-1}
令2(x+3/2)/√5=u，则(2/√5)dx=du，dx=(√5/2)du，代入上式得：
∫(2x+3)/(x²+3x+1)dx=(4/5)∫(√5/2)du/(u²-1)=(2/√5)∫du/(u+1)(u-1)
=(1/√5)∫[1/(u-1)-1/(u+1)]du=(1/√5)[ln∣u-1∣-ln∣u+1∣]+C
=(1/√5)ln∣(u-1)/(u+1)∣+C=(1/√5)ln{∣2(x+3/2)/√5-1]/[2(x+3/2)/√5+1∣}
=(1/√5)ln∣(2x+3-√5)/(2x+3+√5)∣+C..............(2)

(1)和(2)能一样吗？

追问

∫1/(x²+3x+1)dx=∫dx/[(x+3/2)²-5/4]=(4/5)∫dx/{[2(x+3/2)/√5]²-1}这一步可以更具体点吗？

追答

∫1/(x²+3x+1)dx=∫dx/[(x+3/2)²-5/4]=∫dx/{(5/4)[(x+3/2)²/(5/4)-1]}【分母上提出常数5/4】

=(4/5)∫dx/[(x+3/2)²/(5/4)-1]【把提出的5/4颠倒后拿到积分符号外边】
=(4/5)∫dx/{[2(x+3/2)/√5]²-1}【其中(x+3/2)²/(5/4)=4(x+3/2)²/5=[2(x+3/2)/√5]²】

Answer 2

当然不一样
∫(2x+3)/(x^2+3x+1)dx=ln|x^2+3x+1|+C
∫1/(x^2+3x+1)dx=∫1/(x+3/2)^2-5/4)dx
=(1/√5)ln|(x-√5/2)/(x+√5/2)|+C