Problem 29 - Essential Calculus

Question

Verify the integration formula. $$ \int \arctan u d u=u \arctan u-\ln \sqrt{1+u^{2}}+C $$

Step-by-Step Solution

Answer

The given formula is correct. The final integrated formula is \( u \arctan u -\ln|\sqrt{1 + u^{2}}| + C \)

Step 1: Identify the components

Identify the functions 'u' and 'v'. Let's say, 'u' = arctan(u) and 'v' = du.

Step 2: Calculate the integrals and derivatives

Compute the integral(dv) of the 'v', and the derivative(u') of 'u'. So, \(\int dv = \int du = u\) and \(u' = d(\arctan u)/ du = 1 / (1 + u^2)\)

Step 3: Apply the Integration by parts formula

Plug these values into the integration by parts formula. Result is \(u\int v dx - \int u'( \int v dx) dx = u(\int du) - \int (1 / (1 + u^2)) (u du) = u \arctan u - \int (u / (1 + u^2)) du \)

Step 4: Simplify the integral

The integral can be simplified by using the variable substitution method, let \( p = 1 + u^{2} \). So, \( \frac{dp}{du} = 2u \), hence \( du = dp/(2u) \). When you sub these into the integral you get : \( -\int dp/2 = -\ln|p|/2 = -\ln|\sqrt{1 + u^{2}}| \)

Step 5: Add the constant of integration

Finally add the constant of integration, 'C' to derive the result \( u \arctan u -\ln|\sqrt{1 + u^{2}}| + C \)