Question

有没有大神会用第一类换元法，求下列不定积分的。谢谢啊，急求15.16.17.18.

Answer 1

(15)
令x²=t
∫[x/(1+x⁴)]dx
=½∫d(x²)/[1+(x²)²]
=½∫[1/(1+t²)]dt
=½arctant +C
=½arctanx² +C
(16)
令2x=t
∫3^(2x)dx
=½∫3^(2x)d(2x)
=½∫3^tdt
=3^t/(2ln3) +C
=3^(2x)/(2ln3) +C
=9^x/(2ln3) +C
(17)
令lnx=t
∫(lnx/x)dx
=∫lnxd(lnx)
=∫tdt
=½t²+C
=½ln²x +C
(18)
令lnx=t
∫[1/(xlnx)]dx
=∫(1/lnx)d(lnx)
=∫(1/t)dt
=ln|t| +C
=ln|lnx| +C