How to Integrate Ln(x^2 + 1)
The natural logarithm of x^2 + 1 is a function whose graph looks like that of the simpler function y = Ln(x^2) = 2*Ln(x). For large values of x, the functions are practically indistinguishable, however, their behavior near x = 0 and negative numbers is quite divergent. The function y = 2*Ln(x) has a vertical asymptote at x = 0, and is undefined for x = 0 and negative values of x. However, the function y = Ln(x^2 + 1) has no asymptotes and is defined for all values of x. You can compare their graphs in the image below.
The functions y = Ln(x^2 + 1) and y = 2*Ln(x) are also dissimilar in their antiderivatives. The integral of 2*Ln(x) is simply
∫ 2*Ln(x) dx = 2x*Ln(x) - 2x + c
while that of Ln(x^2 + 1) is more complicated to extract. (See linked article for explanation of how to integrate the natural logarithm.) The integral of Ln(x^2 + 1) is also dissimilar to that of another closely related function, y = Ln(x^2 - 1). The integral of Ln(x^2 - 1) is fairly easy to find once you make the transformation Ln(x^2 - 1) = Ln(x-1) + Ln(x+1). This gives you
∫ Ln(x^2 - 1) dx
= ∫ Ln(x-1) dx + ∫ Ln(x+1) dx
= (x-1)*Ln(x-1) - (x-1) + (x+1)*Ln(x+1) - (x+1) + c
= (x-1)*Ln(x-1) + (x+1)*Ln(x+1) - 2x + c
Fortunately, the antiderivative of Ln(x^2 + 1) can be found using the technique of integration by parts, and a little algebraic to convert the resulting integral expression into a familiar integral.
Graphs of 2*Ln(x), Ln(x^2 + 1) and Ln(x^2 - 1)
Integrating Ln(x^2 + 1)
Recall the integration by parts formula,
∫ u * dv = uv - ∫ v * du
To find the antiderivative of Ln(x^2 + 1), we set u = Ln(x^2 + 1) and dv = dx. This produces the related equations
du = 2x/(x^2 + 1) dx
v = x
Now we can transform our original logarithmic function integral into a simpler rational function integral:
∫ Ln(x^2 + 1) dx
= x*Ln(x^2 + 1) - ∫ (2x^2)/(x^2 + 1) dx
To make this integral even simpler, we use a little algebra to come up with an equivalent expression for (2x^2)/(x^2 + 1):
(2x^2)/(x^2 + 1)
= (2x^2 + 2)/(x^2 + 1) - 2/(x^2 + 1)
= 2 - 2/(x^2 + 1)
Plugging this back into the integral equation gives us
∫ Ln(x^2 + 1) dx
= x*Ln(x^2 + 1) - ∫ 2 dx + 2*∫ 1/(x^2 + 1) dx
Since the antiderivative of 1/(x^2 + 1) is the arctangent function, we get
∫ Ln(x^2 + 1) dx
= x*Ln(x^2 + 1) - 2x + 2*arctan(x) + c
As you can see, this antiderivative formula is much more complicated than that of 2*Ln(x) or Ln(x^2 - 1) as it involves not only a natural log function, but also an inverse trig function.
Summary of Integral Formulas
Find the area under the curve y = Ln(x^2 + 1) from x = 0 to x = 4.
Since we know the antiderivative of Ln(x^2 + 1), we simply need to plug the limits of integration into the integral formula and subtract. This gives us
[4*Ln(16+1) - 2*4 + 2*arctan(4)] - [0*Ln(1) - 2*0 + 2*arctan(0)]
Solve the separable differential equation y' = Ln(x^2 + 1)/e^y with the initial condition y(0) = 0
First we rewrite the differential equation as
dy/dx = Ln(x^2 + 1)/e^y
Now we use the separation of variables technique to transform it into an equivalent differential expression:
e^y dy = Ln(x^2 + 1) dx
Next, we integrate both sides:
∫ e^y dy = ∫ Ln(x^2 + 1)
e^y = x*Ln(x^2+1) - 2x + 2*arctan(x) + c
Then we solve for y:
y = Ln[ x*Ln(x^2+1) - 2x + 2*arctan(x) + c ]
And finally, we plug in the initial conditions to solve for c:
0 = Ln[ 0 + c ]
1 = c
Therefore, the complete solution to the differential equation is
y = Ln[ x*Ln(x^2+1) - 2x + 2*arctan(x) + 1 ]
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