求大神帮忙做一下高数题 感谢

求大神帮忙做一下高数题 感谢感谢
2025-05-07 03:10:36
推荐回答(1个)
回答1:

令t=√(x^2-9),t^2=x^2-9,
2tdt=2xdx tdt=xdx
积分号下:√(x^2-9)dx/x
=√(x^2-9) xdx/x^2 (分子分母同乘以x)
=t *tdt/(t^2+9)
=t^2dt/(t^2+9)
=[1-9/(t^2+9)]dt
∫[1-9/(t^2+9)]dt
=t-3arctan(t/3)+C
=√(x^2-9)-3arctan[√(x^2-9)/3]+C