Question Number 212456 by Hanuda354 last updated on 14/Oct/24

Answered by Sutrisno last updated on 14/Oct/24

Commented by Hanuda354 last updated on 14/Oct/24

Answered by Ghisom last updated on 14/Oct/24
![∫ ((√(9x^2 −49))/x^3 )dx= [t=arcsin ((14(√(9x^2 −49)))/x^2 ) → dx=−((x(√(9x^2 −49)))/(14))dt] =−(9/(28))∫(1+cos t)dt=−(9/(28))(t+sin t)= =−((√(9x^2 −49))/(2x^2 ))−(9/(28))arcsin ((14(√(9x^2 −49)))/x^2 ) +C](https://www.tinkutara.com/question/Q212471.png)
Commented by Hanuda354 last updated on 14/Oct/24
