What is the intergal of (1+t^2)^1/2/t?

Question

to rewrite it... squareroot of 1+t^2 divided by t

Answer 1

Try a Trig subtitution.

Answer 2

The answer is: -[(1/t^2)+(1/t^4)]
Work Procedure:
intergal of (1+t^2)^1/2/t = intergal [(sqrt(1+t^2)/t^2] dt
= intergal [sqrt{(1+t^2)/t^2}] dt
= intergal [sqrt(1/t^2 + 1)] dt
= intergal [1/t^2 + 1]^(1/2) dt
= {[(1/t^2 + 1)^3/2 / (1/2)}[-(1/2t^2)]
since integral of x^n = (x^n+1)/n
= -(1/t^2)[1 + 1/t^2]^3/2

Answer 3

Use substitution of t=tanx, dt= secx^2

int sqrt(1+tanx^2)/tanx (secx^2)

1+tanx^s =secx^2

Int sqrt(secx^2)/tanx(secx^2)

Int secx^3/tanx dx

Int 1/cosx^3/(sinx/cosx)

Int sinx/(cosx^4)

U= cosx
Du= -sinx

Int – 1/u^4

1/u^3

1/(cosx^3)

Secx^3