Expert: Scotto - 3/11/2009

Question
How do I know when to use integration by parts? My textbook uses ln to solve the integral of x/square root of (9 - x ^2) but not for 1/ square root of (x + 1)

Answer
The only way that I have found is by trial and error.
Eventually you recognize how to do a lot of them.

On the x/√(a+bx²), it can also be done by u-substitution.
Let u=a+bx², then du = 2bx dx.  
This can replace the x dx by noting that x dx = du /(2b).

On the 1/√(x+1), I would say let u=√(x+1).  
This would make du = dx/(2√(x+1)), so 2 du = dx/√(x+1).
So ∫ 1/√(x+1) dx would become ∫ 2 du.
The answer is just 2u, which is 2√(x+1).

To test this, take the derivative.  
Note that √(x+1) is the same as (x+1)^0.5.
When this is differentiated, the 0.5 cancels the 2 out front.
One is subtracted from the exponet, making the exponet be -0.5.
This makes the answer (x+1)^-0.5 = 1/√(x+1).

There are generally three cases for integration when a power of an expression is in the denominator.

1: The derivative of the denominator is in the numerator
(or a constant time the derivative).  This is u-substitution.

2: There is an x² in the squareroot with no x in the denominator.
This can be converted to a trig function.  If it is a plus sign in the denominator beneath the squareroot, x goes on the far leg and the squareroot of the other term goes on the near leg, making the hypoteneuse into the factor that is in the denominator.  If there is a minus sign in the squareroot, the hypoteneuse is the squareroot of the + term and the far side is the squareroot of the - term.  This makes the near side what you are integrating.

3: Occasionally (that's rarely) they involve integration by parts.
Let dv be the denominator with the derivative on top.  Let u be the rest of the function.  Do a u-v integration.  ∫u dv = uv - ∫v du.
I'm not sure you have seen these.  Suppose the problem were
∫x³/√(x²+1) dx.  For this problem, I might let
u=x² and dv=x/√(x²+1) dx.  That would give us du = 2x dx and
v=√(x²+1).  We would have ∫x³/√(x²+1) dx = uv - ∫v du =
x√(x²+1) - ∫2x√(x²+1)dx.  On the integral, let u=x²+1, then du=2x.
The integral reduces to ∫√u du, which is 2u^(3/2)/3.

Of course, even in this example, you might get away with letting the far leg of a triangle be x, the near leg be 1, and the hypoteneuse be √(x²+1).  We would then have tan(Θ) = x, so sec²(Θ)dΘ = dx and
√(x²+1) would be the sec(Θ).

Substituting this back in to ∫x³/√(x²+1) dx would give us
∫(tan³(Θ)/sec(Θ)) sec²(Θ) dΘ, which would reduce to
∫tan³(Θ)sec(Θ)dΘ.  Now for this problem, we would also need to substiture tan²(x) = sec²(Θ) + 1, so we'd have
∫tan(Θ)sec³(Θ) + tan(Θ)sec(Θ) dΘ.
Now to integrate that, note that the derivative of sec(Θ) is tan(Θ)sec(Θ), so on the first one do a u-substitution and get ∫u²du.  On the second one, let u=sec(Θ), so du=sec(Θ)tan(Θ)d Θ, and you are just doing ∫1 du.

Again, there are many forms and how to solve them is by trial and error.