Mathematics

Question

integrate by substitution
a) ∫(√1+x²) x⁵ dx

Answer 1

∫√(1+x²)x⁵dx = ∫√(1+x²)x⁴xdx

 

    Let u = 1+x²   So, du = 2xdx

 

                               xdx = (1/2)du  and x⁴ = (x²)² = (u-1)²

 

Therefore, the given integral can be rewritten as:

 

   ∫(√u)(u-1)²(1/2)du = (1/2)∫(√u)(u²-2u+1)du

 

                                 = (1/2)∫[u^5/2 - 2u^3/2 + u^1/2]du

 

                                 = (1/2)[(2/7)u^7/2 - (4/5)u^5/2 + (2/3)u^3/2] + C

 

= (1/7)(1+x²)^7/2 - (2/5)(1+x²)^5/2 + (1/3)(1+x²)^3/2 + C

 

Answer 2

I will try to get you started: draw a rt. triangle with acute angle θ having opposite leg 1, adjacent leg, so hypotenuse is √(1+x²).  NOTE: I have assumed that the parenthesization should be ∫√(1+x²) x⁵ dx.

 

Under these circumstances dx = -1/sin²θ = - csc²θ dθ and the integral can be rewritten as 

∫cscθ cot⁵θ (-csc²θ) dθ.

 

Can you proceed from here(if you agree that this is a correct approach)>