How do you evaluate the integral #int ln(1+x^2)#?

Question

Luke Alvarez · Accepted Answer

#intln(1+x^2)dx=xln(1+x^2)-2x+2arctan(x)+C#

#I=intln(1+x^2)dx#

Integration by parts takes the form #intudv=uv-intvdu#. For the given integral, let:

#u=ln(1+x^2)#

#color(white)(u=ln(1+x^2))du=(2x)/(1+x^2)dx#

#dv=dx#

#color(white)(dv=dx)v=x#

Then:

#I=uv-intvdu#

#I=xln(1+x^2)-int(2x^2)/(1+x^2)dx#

#I=xln(1+x^2)-int(2(1+x^2)-2)/(1+x^2)dx#

#I=xln(1+x^2)-int(2-2/(1+x^2))dx#

#I=xln(1+x^2)-2intdx+2intdx/(1+x^2)#

#I=xln(1+x^2)-2x+2arctan(x)+C#

Christopher Allers · Answer

undefined To evaluate the integral $\int \ln(1 + x^2) \, dx$, we can use integration by parts. Let $u = \ln(1 + x^2)$ and $dv = dx$, then $du = \frac{2x}{1 + x^2} \, dx$ and $v = x$.

Applying the integration by parts formula:

$$
\int u \, dv = uv - \int v \, du
$$

Substitute the values:

$$
\int \ln(1 + x^2) \, dx = x \ln(1 + x^2) - \int \frac{2x^2}{1 + x^2} \, dx
$$

Now, the integral on the right-hand side is a rational function, which can be integrated using partial fraction decomposition or trigonometric substitution, depending on the preferred method.

After integrating the second term, you will have the final result for the integral $\int \ln(1 + x^2) \, dx$.