Integration

Question

How to integrate following

 

∫x(x+1)^1/2 dx

 

∫2x(x²+1)dx

Answer 1

∫x(x+1)^½dx   Let u = x+1  Then, du=dx and x = u-1

 

So, the given integral is equivalent to ∫(u-1)u^½du

 

                                                  = ∫[u^3/2 - u^1/2]du

 

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

 

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

---------------------------------------------------------------------------------

∫2x(x²+1)dx = ∫(2x³+2x)dx = 2(x⁴/4) + x² + C

 

                                       = (1/2)x⁴ + x² + C

 

Answer 2

The fun part about integration is there are a variety of tools available, so it's a matter of creativity to pick ones that work.  For the first, I would note that (x+1)^1/2 is easy to integrate and x disappears when you differentiate, so I would apply integration by parts.  Let u = x and dv = (x+1)^1/2dx.  Then, 

 

∫x(x+1)^1/2dx = ∫udv = uv - ∫vdu 

= x * (x+1)^3/2 / (3/2) - ∫[(x+1)^3/2 / (3/2)]dx

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

 

You can simplify that final expression quite a bit, but I'll leave that for you.

 

For the second, we could just multiply it out and use power rule on each term (actually the easiest approach).  But, perhaps it's a bit more elegant to make a substitution u = x² + 1.  Then du = 2x dx.  Our integral becomes:

 

∫u du = u² / 2 + C = (x² +1)² / 2 + C

= x⁴ / 2 + x² + 1/2 + C

= x⁴ / 2 + x² + C

 

If you like, you can confirm that this gives the same answer as if we multiplied it out and applied power rule directly.

 

If you have additional questions on antiderivatives, feel free to ask.