Problem 23 - Essential Calculus

Question

Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int \arctan x d x $$

Step-by-Step Solution

Answer

The integral \( \int \arctan(x) dx = \arctan(x)*x - \frac{1}{2} \ln |1 + x^2| + c \)

Step 1: Identify u and dv

To use integration by parts, we first need to choose two parts of the integral, i.e. \( u \) and \( dv \). Here, let's identify \( u = \arctan(x) \) and \( dv = dx \).

Step 2: Compute du and v

Calculate \( du \) and \( v \) using \( du = \frac{1}{1 + x^2} dx \) (as the derivative of \( \arctan(x) = \frac{1}{1+ x^2} \)), and \( v = \int dx = x \).

Step 3: Apply Formula for Integration by parts

The formula for integration by parts is \( \int u dv = u*v - \int v du \). Applying this formula we get: \( \int \arctan(x) dx = \arctan(x)*x - \int x*\frac{1}{1 + x^2} dx \)

Step 4: Simplify the integral

We see that the new integral we need to calculate is simpler: \( \int x*\frac{1}{1 + x^2} dx \). This integral can be solved by substituting the denominator \( 1+x^2 = z \). Therefore \( dz = 2x dx \) or \( dx = \frac{dz}{2x} \). Replacing this in our integral gives us \( \frac{1}{2} \int \frac{dz}{z} = \frac{1}{2} \ln |1 + x^2| + c \). The final answer combines both parts: \( \arctan(x)*x - \frac{1}{2} \ln |1 + x^2| + c \)