求√(1+t²)的原函數(shù)
即∫√(1+t²)dt=√(1+t²)*t-∫td√(1+t²)=√(1+t²) *t-∫t²/√(1+t²)dt
=√(1+t²)*t-∫[(√(1+t²))²-1]/√(1+t²)dt
=√(1+t²)*t-∫√(1+t²)dt+∫1/√(1+t²)dt 此步化下一步的證明在最下方
=√(1+t²)*t-∫√(1+t²)dt+ln(t+√(1+t²))+C
所以∫√(1+t²)dt=1/2√(1+t²)*t+1/2ln(t+√(1+t²))+C2 C2=1/2)*C
求∫1/√(a²+x²) dx(a>0)
令x=atant,t∈(-π/2,π/2),dx=asec²tdt,√(a²+x²)=asect.
帶入所求積分得
∫1/√(a²+x²) dx=∫(asec²t)/(asect) dt=∫sect dt=ln|sect+tan t|+C
.因?yàn)閠ant=x/a,所以sect=√(x²+a²)/a.
因此 ∫1/√(a²+x²)dx=ln(x/a+(√(x²+a²))/a)+C1
=ln(x+√(x²+a²))+C
其中C=C1-lna