How do you use substitution to integrate x(sqrt(2x+1))dx.?

2 Answers
Jun 10, 2015

This is better-solved using integration by parts. There is nothing u-substitution-related that is easy to think of quickly, here.

I would choose the function that is easier to differentiate and make go away as u, such as x^n.

Let:

  • u = x
  • du = dx
  • dv = (2x+1)^(1//2)dx
  • v = (2//3)/2(2x+1)^(3//2) = 1/3(2x+1)^(3//2)

Then, we get:

int udv = uv - intvdu

= x/3(2x+1)^(3//2) - int1/3(2x+1)^(3//2)dx

= x/3(2x+1)^(3//2) - 1/3(1/5(2x+1)^(5//2))

= x/3(2x+1)^(3//2) - 1/15(2x+1)^(5//2)

= (2x+1)^(3//2)[x/3 - 1/15(2x+1)]

= 1/15(2x+1)^(3//2)[5x - (2x+1)]

And now we tack on the constant at the end.

= color(blue)(ul(1/15(2x+1)^(3//2)(3x-1) + C))

Aug 1, 2017

Please see below.

Explanation:

int xsqrt(2x+1) dx

Let u be the radicand (the stuff under the radical).

So u = 2x+1, then x = 1/2(u-1) and du = 2 dx

int xsqrt(2x+1) dx = 1/2 int x(2x+1)^(1/2) 2 dx

= 1/2 int 1/2(u-1)u^(1/2)du

= 1/4 int (u^(3/2)-u^(1/2)) du

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

= 1/10u^(5/2)-1/6u^(3/2)+C

Simplify as desired before reversing the substitution.

I like

sqrtu/30 u(3u-5) +C

So we have

int xsqrt(2x+1) dx = sqrt(2x+1)/30 (2x+1) (3(2x+1)-5) +C

= sqrt(2x+1)/30 (2x+1) (6x-2) +C

= sqrt(2x+1)/15 (2x+1) (3x-1) +C