let x=atanu dx=a(secu)^2 du ∫ dx/(x^2+a^2)^(3/2) =∫ a(secu)^2 du/[ a^3. (secu)^3]
=(1/a^2)∫ (cosu)^2 du =[1/(2a^2)]∫ (1+cos2u) du =[1/(2a^2)] [u+(1/2)sin2u] +C =[1/(2a^2)] [arctanu(x/a)+ ax/(x^2+a^2)] +C (2) let x=secu dx=secu.tanu du ∫ dx/[x+√(x^2-1)] =∫ [x-√(x^2-1)] dx =(1/2)x^2 -∫ √(x^2-1) dx =(1/2)x^2 -∫ secu.(tanu)^2 du =(1/2)x^2 -∫ secu.[(secu)^2-1] du =(1/2)x^2 +ln|secu+tanu| -∫ (secu)^2 du =(1/2)x^2 + ln|secu+tanu| -(1/2)[secu.tanu +ln|secu+tanu|] +C =(1/2)x^2 + (1/2)ln|secu+tanu| -(1/2)secu.tanu +C =(1/2)x^2 + (1/2)ln|x+√(x^2-1)| -(1/2)x.√(x^2-1) +C
// ∫ (secu)^3 du =∫ secudtanu =secu.tanu -∫ (secu).(tanu)^2 du =secu.tanu -∫ (secu).[(secu)^2-1] du 2∫ (secu)^3 du =secu.tanu +∫ secu du ∫ (secu)^3 du =(1/2)[secu.tanu +ln|secu+tanu|] +C'