Ta có:
2x2+1√x2+2x=2x2+1√(x+1)2−1=a[(x+1)2−1]√(x+1)2−1+b(x+1)√(x+1)2−1+c√(x+1)2−1
=ax2+(2a+b)x+b+c√x2+2x
Đồng nhất đẳng thức, ta được:
{a=22a+b=0b+c=1↔{a=2b=−4c=5
Khi đó:
2x2+1√x2+2x=2√(x+1)2−1−4(x+1)√(x+1)2−1+5√(x+1)2−1
Do đó:
∫f(x)dx=∫[2√(x+1)2−1−4(x+1)√(x+1)2−1+5√(x+1)2−1]dx
=(x+1)√x2+2x−ln|x+1+√x2+2x|−4√x2+2x+5ln|x+1
+√x2+2x|+C
=(x+1)√x2+2x+4ln|x+1+√x2+2x|−4√x2+2x+C