∫ dx / ((2x + 1)2 – √(2x + 1))
(2x+1)=t6
Тогда
∛(2x+1)2=t4
√2x+1=t3
x=(t6–1)/2
dx=6t5dt/2
dx=3t5dt
получаем
∫ 3t5dt/(t4–t3)= 3· ∫ t2dt/(t–1)= 3· ∫ (t2–1+1)dt/(t–1)=
=3· ∫ (t2–1)dt/(t–1) + 3·∫(dt/(t–1))=
=3· ∫ (t+1)dt +3 ∫ dt/(t–1)=
=3(t2/2)+3t + 3ln|t–1| + C
Обратный переход
= (3/2)∛(2x+1)+3(2x+1)1/6 – 3·ln|(2x+1)1/6 – 1| + C