Question

高数求不定积分。

Answer 1

如图所示，用换慧岁宏元雀谨来做前册，注意三角函数的变形

Answer 2

=∫(1-√(x²-1))/(2-x²)dx
=1/2√2∫穗察1/(√2+x)+1/(√2-x)dx-∫tanu/(2-sec²u)dsecu
=(ln|√2+x|-ln|√2-x|)/2√2+∫(sec³u-secu)/(sec²u-2)du
积分部分=∫伏厅secu+secu/(sec²u-2)du
=ln|secu+tanu|+1/2√2∫secu/(secu-√2)-secu/(secu+√2)du
其中
∫secu/(secu-√2)du
=∫1/(1-√2(2cos²(u/2-1))du
=∫sec²(u/2)/((1+√2)sec²(u/2)-2√2)du
=2∫1/((1+√2)tan²(u/2)+(1-√缺族隐2))dtan(u/2)
=2∫1/((√2+1)²tan²(u/2)-1)d(√2+1)tan(u/2)
=ln|(√2+1)tan(u/2)-1|-ln|(√2+1)tan(u/2)+1|
而
∫secu/(secu+√2)du
=∫sec²(u/2)/((1-√2)sec²(u/2)+2√2)du
=2∫1/((1-√2)tan²(u/2)+1+√2)dtan(u/2)
=2∫1/(1-(√2-1)²tan(u/2))d(√2-1)tan(u/2)
=ln|1+(√2-1)tan(u/2)|-ln|1-(√2-1)tan(u/2)|