solving integrals

Question

∫ ( (dx)/√(x²+36) )=_____________

 

 

a. ( (√(x²+36)+x)/6 )+C

 

b. ln abs( (x+6)/(√(x2+36)) )+C

 

c. ln ( (√(x2+36)+x)/6 )+C

 

 

 

Answer 1

A standard integral shown in tables of integrals is:

 

∫(1/sqrt(x²+a²))dx=Ln[x+sqrt(x²+a²)]

 

In this case, a²=36

 

Choice (c) has this form.

 

The integral can be restated so as to give the form in choice (c), so that is your choice.

 

∫(1/sqrt(x²+a²))dx=∫(1/sqrt((x/a)²+1))(dx/a)

 

define w=x/a

 

dw=dx/a

 

∫(1/sqrt((x/a)²+1))(dx/a)=∫(1/sqrt(w²+1))dw

 

This is the integral from the table, above, with a=1, so

 

∫(1/sqrt(w²+1))dw=LnLn[w+sqrt(w²+1)]=Ln[(x/a)+sqrt((x/a)²+1)]

 

Ln[(x/a)+sqrt((x/a)²+1)]=Ln[(x/a)+(sqrt(x²+a²))/a]= choice (c)

 

 

 

 

Answer 2

Let x = 6tanθ, then tanθ = x/6 and dx = 6sec²θdθ

 

So, √(x²+36) = √(36tan² θ + 36) = √[36(tan²θ+1)]

 

                                                           = √(36sec²θ)

 

                                                           = 6secθ

 

The original integral is equivalent to 

 

  ∫[(6sec²θ)/(6sec θ)]dθ

 

            = ∫secθdθ

 

            = ln l secθ+tanθ l+C

 

            = ln l (√(x²+36)/6 + x/6) l + C

 

            = ln l (√(x²+36) + x)/6 l + C