Question

1/(1+t^2)^2的积分是什么

求不定积分

Answer 1

∫[1/(1+t²)²]dt，令t=tanu,dt=sec²udu

=∫[sec²u/(1+tan²u)²]du

=∫(sec²u/sec^4u)du

=∫(1/sec²u)du

=∫cos²udu

=(1/2)∫(1+cos2u)du

=(1/2)∫du+(1/2)(1/2)∫cos2ud(2u)

=(1/2)u+(1/4)sin2u+C

=(1/2)u+(1/4)*2sinucosu+C

=(1/2)arctant+(1/2)[t/√(1+t²)][1/√(1+t²)]+C

=(1/2)arctant+(1/2)[t/(1+t²)]+C

=(1/2)arctant+t/[2(1+t²)]+C

不定积分的公式：

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

Answer 2

∫[1/(1+t²)²]dt，令t=tanu,dt=sec²udu
=∫[sec²u/(1+tan²u)²]du
=∫(sec²u/sec^4u)du
=∫(1/sec²u)du
=∫cos²udu
=(1/2)∫(1+cos2u)du
=(1/2)∫du+(1/2)(1/2)∫cos2ud(2u)
=(1/2)u+(1/4)sin2u+C
=(1/2)u+(1/4)*2sinucosu+C
=(1/2)arctant+(1/2)[t/√(1+t²)][1/√(1+t²)]+C
=(1/2)arctant+(1/2)[t/(1+t²)]+C
=(1/2)arctant+t/[2(1+t²)]+C