Question

∫3x²-x+4/(x-1)(2+x²)dx

∫【（x+1）（x^2-2x-1）^1/2】dx
答案是【1/3（x^+x+4）（x^2-2x-1）^1/2】-2ln【x-1+（x^2-2x-1）^1/2】+c

Answer 1

∫(x+1)√(x²-2x-1)dx
=∫(x+1)√[(x-1)²-2]dx,设u=x-1,x=1+u,dx=du
=∫(1+u+1)√(u²-2)du
=∫(2+u)√(u²-2)du,设u=√2secP,du=√2tanPsecPdP
=∫[(2+√2secP)√(2sec²P-2)*(√2tanPsecP)]dP
=∫[(2+√2secP)√2tanP*(√2tanPsecP)]dP
=2∫[(2+√2secP)tan²PsecP]dP
=4∫tan²PsecPdP+2√2∫tan²Psec²PdP
=4∫[(sec²P-1)secP]dP+2√2∫tan²Pd(tanP)
=2√2*tan³P/3+4∫sec³PdP-4∫secPdP
=(2√2/3)tan³P+4(1/2*sinPsec²P+1/2*∫secPdP)-4∫secPdP
=(2√2/3)tan³P+2tanPsecP-2∫secPdP
=(2√2/3)tan³P+2tanPsecP-2ln|secP+tanP|+C
代回P=arcsec(u/√2),tanP=√(u²-2)/√2,secP=u/√2
=(2√2/3)*[√(u²-2)/√2]³+2[√(u²-2)/√2](u/√2)-2ln|√(u²-2)/√2+u/√2|+C
=2√2/3*(u²-2)√(u²-2)/(2√2)+u√(u²-2)-2ln|[√(u²-2)+u]/√2|+C
=1/3*(u²-2)√(u²-2)+u√(u²-2)-2[ln|√(u²-2)+u|-ln√2]+C
=1/3*√(u²-2)*[(u²-2)+3u])-2ln|√(u²-2)+u|+ln2+C
代回u=x-1
=1/3*√(x²-2x+1-2)*(x²-2x+1-2+3x-3)-2ln|√(x²-2x+1-2)+x-1|+ln2+C
=(1/3)(x²+x-4)√(x²-2x-1)-2ln|x-1+√(x²-2x-1|+ln2+C
=(1/3)(x²+x-4)√(x²-2x-1)-2ln|2[x-1+√(x²-2x-1)]|+C