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Chapter 1: Problem 14

Deduce the reduction formula for \(I_{n}=\int \frac{d x}{\left(1+x^{4}\right)^{n}}\) andhenceevaluate \(I_{2}=\int \frac{d x}{\left(1+x^{4}\right)^{2}} .\)

Short Answer

Expert verified
The reduction formula for the given integral is $I_{n} = \int\frac{1}{\left( 1+x^{4}\right)^{n-1}} dx-\int\frac{I_{n-1}}{1+x^{4}} dx$.

Step by step solution

01

Choose u and dv/dx

Let's choose \(u = x\) and \(\frac{dv}{dx} =\frac{1}{\left( 1+x^{4}\right)^{n}}\).
02

Compute du/dx and v

Differentiate u to find \(\frac{du}{dx}\) and integrate \(\frac{dv}{dx}\) to find \(v\). \(\frac{d u}{d x}=1\) and \(v = \int \frac{1}{\left( 1+x^{4}\right)^{n}} dx = I_{n}\).
03

Apply the integration by parts formula

Plugging in \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the integration by parts formula, we get: \(I_{n}=\int \frac{d x}{\left(1+x^{4}\right)^{n}} = u v - \int v \frac{d u}{d x} d x = xI_{n}-\int I_{n}dx\)
04

Solve for the reduction formula

Rearrange the previous equation for \(I_{n}\): \(I_{n}+\int I_{n} dx= xI_{n}\) Divide the equation by \(1+x^{4}\): \(\int\frac{1}{\left( 1+x^{4}\right)^{n}} dx+\int\frac{I_{n}}{1+x^{4}} dx= xI_{n}\) Now we have derived the reduction formula: \(I_{n} = \int\frac{1}{\left( 1+x^{4}\right)^{n-1}} dx-\int\frac{I_{n-1}}{1+x^{4}} dx\)
05

Apply the reduction formula for \(n=2\)鈥�

To find the integral for \(n=2\), let's use the reduction formula we derived: \(I_2 = \int\frac{1}{\left( 1+x^{4}\right)^{1}} dx - \int\frac{I_1}{1+x^{4}} dx\) Since \(I_1 = \arctan(x^2) + C\), the integral becomes: \(I_2 = \int\frac{1}{ 1+x^{4}} dx - \int\frac{\arctan(x^2)}{1+x^{4}} dx\) This integral does not have a closed-form solution, and in order to evaluate it, we would need to use numerical methods or definite integral with given bounds.

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