Question

高数题 用积分的换元法与分部积分法计算下列定积分 第6题和第10题 急

Answer 1

Answer 2

（6）原式=(1/2)*∫(2x+3-3)/(x^2+3x+2)dx
=(1/2)*∫(2x+3)/(x^2+3x+2)dx-(3/2)*∫dx/(x+1)(x+2)
=(1/2)*∫d(x^2+3x+2)/(x^2+3x+2)-(3/2)*∫[1/(x+1)-1/(x+2)]dx
=(1/2)*ln|x^2+3x+2|-(3/2)*[ln|x+1|-ln|x+2|]+C，其中C是任意常数

（10）令x=√2sint，则dx=√2costdt
原式=∫(0,π/2) √2cost*√2costdt
=∫(0,π/2) 2cos^2tdt
=∫(0,π/2) (1+cos2t)dt
=[t+(1/2)*sin2t]|(0,π/2)
=π/2