Question

求4∫1/(t^2-1)dt解的具体步骤，谢谢！

另外4∫1/(2+u^2)dt=2*根二arctan u/根二+C，怎么解出来的？

Answer 1

4∫ 1/(t²-1) dt
= -4∫ 1/(1-t²) dt
令t = sinz，dt = cosz dz
tanz = t/√(1-t²)，secz = 1/√(1-t²)
原式= -4∫ cosz / cos²z dz
= -4∫ secz dz
= -4ln|secz + tanz| + c
= -4ln|(1+t) / √(1-t²)| + c
= -4ln|(1+t) / √(1-t)√(1+t)| + c
= -2ln| (1+t) / (1-t) | + c

4∫ 1/(2+u²) du
令u = √2*tany，du=√2*sec²y dy
原式= 4∫ √2*sec²y / 2+2tan²y dy
= 2√2*∫ sec²y / sec²y dy
= 2√2 * ∫ dy
= 2√2 * y + c
= 2√2 * arctan(u/√2) + c

It's pity that nobody answers these questions.