Question

(x-2)×x²-3²/x²-8+2x

求不定积分∫
∫dx/[x^2√(1+x^2)
∫(x^3-6x+8)dx/(x-2)
∫dx/√(x^2+6x-16)
3Q

Answer 1

【1】∫dx/[x²√(1+x²)],令x=tanλ,dx=sec²λ
=∫sec²λ/[tan²λ√(1+tan²λ)]dλ
=∫sec²λ/(tan²λsecλ)dλ
=∫cotλcscλdλ
=-cscλ+C
=-(1/x)[√(x²+1)]+C
【2】∫(x³-6x+8)dx/(x-2)
=∫[(x³-8+16)/(x-2)-6x/(x-2)]dx
=∫[(x-2)(x²+2x+4)]dx/(x-2)+16∫dx/(x-2)-6[∫dx+2∫dx/(x-2)]
=∫(x²+2x+4)dx+16∫dx/(x-2)-6∫dx-12∫dx/(x-2)
=∫x²dx+2∫xdx-2∫dx+4∫dx/(x-2)
=(1/3)x³+x²-2x+4ln|x-2|+C
【3】∫dx/√(x²+6x-16)
=∫dx/√(x²+6x+9-25)
=∫dx/√[(x+3)²-25],令s=5secu,ds=5tanusecudu
=∫5tanusecudu/√(25sec²u-25)
=∫5tanusecudu/5tanu
=∫secudu
=ln|tanu+secu|+C
=ln|√(s²-5²)/5+s/5|+C
=ln|√[(x+3)²-25]/5+(x+3)/5|+C
=ln|(1/5)[√(x²+6x-16)+x+3]|+C