Integral [(1+x^2)^1/2]dx=?

Answer 1

Integral [(1+x^2)^1/2]dx
= x(1+x^2)^1/2 - Integral[x^2/(1+x^2)^1/2]dx, integration by part
= x(1+x^2)^1/2- Integral (-1+1+x^2)/(1+x^2)^1/2 dx, add/subtract 1
= x(1+x^2)^1/2+ln[x+sqrt(1+x^2)]
-Integral [(1+x^2)^1/2]dx


Collect the same term Integral [(1+x^2)^1/2]dx to the left side, and solve for Integral [(1+x^2)^1/2]dx,

Integral [(1+x^2)^1/2]dx
= (1/2)[x(1+x^2)^1/2
+ln[x+sqrt(1+x^2)]]

Answer 2

put x = tan(y)

1+x² = 1+tan²(y) = sec²(y)

dx = d(tan(y)) = sec²(y)*dy

∫ [(1+x²)^1/2]dx = ∫ sec(y)*sec²(y)*dy

= ∫ sec(y)*sec²(y)*dy

= ∫ sec³(y)*dy