Integration of a complicated radical function

[kashan123999]

Homework Statement

∫(√x).[√(x+1)] dx

Homework Equations

The Attempt at a Solution

Sorry but i couldn't get it any far than this u substitutionu = √(x+1)
du = [(1)/{2√(x+1)}]dx

→ dx = [2√(x+1)]du

putting in the main expression∫ √(u² - 1). 2u² du

 

[Mark44]

Homework Statement

∫(√x).[√(x+1)] dx

Homework Equations

The Attempt at a Solution

Sorry but i couldn't get it any far than this u substitution


u = √(x+1)
du = [(1)/{2√(x+1)}]dx

→ dx = [2√(x+1)]du

putting in the main expression


∫ √(u² - 1). 2u² du

Here's a different approach. Combine the two radicals like so:
$\int \sqrt{x}\sqrt{x + 1}dx = \int \sqrt{x^2 + x}dx$

Now, complete the square inside the radical to get √[(x + a)² - b²]. At this point a trig substitution should work, assuming you have learned that technique.

 

[kashan123999]

so that was mathematically valid : O to multiply..poor algebra me :(