Question

求积分1。∫[（sinx-cosx)/(sinx+cosx)] dx 2.∫dx/(x²-7x+12)

Answer 1

①=∫[(sinx-cosx)/(sinx+cosx)] dx
=-∫[1/(sinx+cosx)] d(cosx+sinx)
=-ln(sinx+cosx)+C
②②∫dx/(x²-7x+12)
=（1/3)∫{1/(x-5)-1/(x-2)}dx
=(1/3)∫d(x-5)/(x-5)-∫d(x-2)/(x-2)

=(1/3){ln(x-5)-ln(x-2)+C

Answer 2

第一题：-(-sinx+cosx)dx=-d(cosx+sinx)所以
∫[（sinx-cosx)/(sinx+cosx)] dx=-∫[1/(sinx+cosx)] d(cosx+sinx)=-ln(sinx+cosx)+c
第二题：1/(x²-7x+12)=1/{(x-2)(x-5)}=(1/3){1/(x-5)-1/(x-2)}所以
∫dx/(x²-7x+12)=（1/3）∫{1/(x-5)-1/(x-2)}dx=(1/3)∫d(x-5)/(x-5)-∫d(x-2)/(x-2)
=(1/3){ln(x-5)-ln(x-2)+c}