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Chapter 6: Problem 29

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

Short Answer

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

Step by step solution

01

Identify the components

Identify the functions 'u' and 'v'. Let's say, 'u' = arctan(u) and 'v' = du.
02

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 \(u' = d(\arctan u)/ du = 1 / (1 + u^2)\)
03

Apply the Integration by parts formula

Plug these values into the integration by parts formula. Result is \(u\int v dx - \int u'( \int v dx) dx = u(\int du) - \int (1 / (1 + u^2)) (u du) = u \arctan u - \int (u / (1 + u^2)) du \)
04

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 : \( -\int dp/2 = -\ln|p|/2 = -\ln|\sqrt{1 + u^{2}}| \)
05

Add the constant of integration

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

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Most popular questions from this chapter

Rewrite the improper integral as a proper integral using the given \(u\) -substitution. Then use the Trapezoidal Rule with \(n=5\) to approximate the integral. $$ \int_{0}^{1} \frac{\cos x}{\sqrt{1-x}} d x, u=\sqrt{1-x} $$

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Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=t^{2} $$
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