Question

急啊，这四题怎么做啊😚

Answer 1

1.

(1)

∫(2x+3)/(x²-7x+10)dx

=⅓∫[13/(x-5) -7/(x-2)]dx

=⅓[13ln|x-5|-7ln|x-2|] +C

(2)

∫1/(x³-1)dx

=(1/6)∫[2/(x-1) -(2x+1)/(x²+x+1) -3/(x²+x+1)]dx

=⅓∫1/(x-1)dx -(1/6)∫(2x+1)/(x²+x+1)dx -(√3/3)∫[1/[1+(2x/√3 + 1/√3)²]d(2x/√3)

=⅓ln|x-1| -(1/6)ln|x²+x+1| -(√3/3)arcan[⅓(2√3x + √3)] +C

2.

(1)

令√(x+1)=t，则x=t²-1

∫[√(x+1)-1]/[√(x+1)+1]dx

=∫(t-1)/(t+1)d(t²-1)

=∫2t(t-1)/(t+1)dt

=∫(2t²+2t-4t-4+4)/(t+1)dt

=∫[2t -4 +4/(t+1)]dt

=t²-4t+4ln|t+1| +C₁

=x+1-4√(x+1) +4ln|√(x+1)+1| +C₁

=x-4√(x+1) +4ln|√(x+1)+1| +C

(2)

令⁶√x=t，则√x=t³，³√x=t²，x=t⁶

∫³√x/[x(√x+³√x)]dx

=∫t²/[t⁶·(t³+t²)]d(t⁶)

=6∫1/[(t²+t)]dt

=6∫[1/t -1/(t+1)]dt

=6(ln|t|-ln|t+1|) +C

=6(ln|⁶√x|-ln|⁶√x +1|) +C