Question

一些高等数学积分问题请高手解决！最好步骤详细一点！ 怕看不懂！谢谢

Answer 1

(1)。∫[r²/(r+4)]dr=∫[r-4+16/(r+4)]dr=∫(r-4)dr+16∫dr/(r+4)
=∫(r-4)d(r-4)+16∫d(r+4)/(r+4)=(r-4)²/2+16ln∣r+4∣+C
(2)。∫dt/(t+4)(t-1)=(1/5)∫[1/(t-1)-1/(t+4)]dt=(1/5)[∫d(t-1)/(t-1)-d(t+4)/(t+4)]
=(1/5)[ln∣t-1∣-ln∣t+4∣]+C=(1/5)ln∣(t-1)/(t+4)∣+C
(3)。【0，1】∫[(x³-4x-10)/(x²-x-6)]dx=【0，1】∫[(x+1)+(3x-4)/(x²-x-6)]dx
=【0，1】{∫(x+1)d(x+1)+∫[(2x-1)/(x²-x-6)+(x-3)/(x²-x-6)]dx}
=【0，1】{∫(x+1)d(x+1)+∫[(2x-1)/(x²-x-6)]dx+∫[1/(x+2)]dx}
=【0，1】{∫(x+1)d(x+1)+∫[d(x²-x-6)/(x²-x-6)]+∫[1/(x+2)]dx}
={(x+1)²/2+ln∣x²-x-6∣+ln∣x+2∣}【0，1】
=(2+ln6+ln3)-(1/2+ln6+ln2)=(3/2)ln(3/2)
(4)。∫[x²/(x-3)(x+2)²]dx=∫{(9/25)/(x-3)+[(16/25)x+(12/25)]/(x+2)²}dx
=(9/25)ln∣x-3∣+(8/25)x²-12/[25(x+2)]+C
(5)。【0，1】∫[x/(x²+4x+13)]dx=【0，1】∫{x/[(x+2)²+9]}dx
=【0，1】{(1/2)∫d[(x+2)²+9]/[(x+2)²+9]-2∫dx/[(x+2)²+9]}
={(1/2)ln(x²+4x+13)-2arctan[(x+2)/3]}【0，1】
=[(1/2)ln18-π/2]-[13/2-2arctan(2/3)]=ln3+(1/2)ln2-(π/2)-(13/2)+2arctan(2/3)

Answer 2

r^2/(r+4)=r-4+16/(r+4)，然后分成两项分别求积分,(r-4)dr和(16/(r+4))dr，就很好算了。总之呢，就是将等式化简，化成最常见的式子然后进行积分啦。