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Chapter 1: Problem 3

Integrate \(\frac{x^{2}}{\left(x^{2}+1\right)^{2}}\) by the substitutions (a) \(x=\tan \theta\) (b) \(\mathrm{u}=\mathrm{x}^{2}+1\), and verify the argeement of the results.

Short Answer

Expert verified
Question: Find the integral of the given expression using two different methods of substitution and verify if the results agree: $\int \frac{x^2}{(x^2 + 1)^2} dx$ Answer: After applying two different substitution methods (Method (a): $x = \tan \theta$ and Method (b): $u = x^2 + 1$), both results agree: $\int \frac{x^2}{(x^2 + 1)^2} dx = x-\arctan(x)+C$.

Step by step solution

01

Express dx in terms of d胃

We are given the substitution \(x = \tan \theta\), we need to find the derivative of \(x\) with respect to \(\theta\) to express \(dx\) in terms of \(d\theta\). Differentiating both sides, \(\frac{dx}{d\theta} = \frac{d(\tan \theta)}{d\theta} = \sec^2 \theta\) So, \(dx = \sec^2 \theta d\theta\)
02

Substitute x and dx

We substitute \(x = \tan \theta\) and \(dx = \sec^2 \theta d\theta\) into the given expression: \(\int \frac{x^2}{(x^2 + 1)^2} dx = \int \frac{\tan^2 \theta}{(\tan^2 \theta + 1)^2} (\sec^2 \theta d\theta)\)
03

Simplify the expression

Recall that the identity \(\sec^2\theta=\tan^2\theta+1\), so the expression can be simplified into \(\int \frac{\tan^2 \theta}{(\sec^2\theta)^2} \sec^2\theta d\theta = \int \frac{\tan^2 \theta}{\sec^4\theta} \sec^2\theta d\theta = \int \frac{\tan^2 \theta}{\sec^2\theta} d\theta\) Now, substitute back \(\sec^2\theta\) with \(\tan^2\theta+1\), and we have \(\int \frac{\tan^2\theta}{\tan^2\theta + 1} d\theta\)
04

Calculating the integral and backward substitution

Now we compute the integral: Let \(v=\tan\theta\), then \(dv=d\theta.\) Thus, \(\int \frac{v^{2}}{v^{2}+1} dv\) Using partial fraction decomposition, we can rewrite that as \(\int \frac{v^{2}+1-1}{v^{2}+1} dv\) \(=\int (1-\frac{1}{v^{2}+1}) dv\) \(=v-\arctan(v)+C\) Now we substitute \(v\) back with \(\tan\theta\) and then substitute \(\tan\theta\) back with \(x\): \(=\tan\theta-\arctan(\tan\theta)+C\) \(=x-\arctan(x)+C\) Method (b): Substituting \(u = x^2 + 1\)
05

Express dx in terms of du

We are given the substitution \(u = x^2 + 1\), we need to find the derivative of \(u\) with respect to \(x\) to express \(dx\) in terms of \(du\). Differentiating both sides, \(\frac{du}{dx} = \frac{d(x^2 + 1)}{dx} = 2x\) So, \(dx = \frac{1}{2x} du\)
06

Substitute x and dx

We substitute \(u = x^2 + 1\) and \(dx = \frac{1}{2x} du\) into the given expression: \(\int \frac{x^2}{(x^2 + 1)^2} dx = \int \frac{u-1}{u^2} (\frac{1}{2x} du)\)
07

Get x in terms of u

To have an integral purely in terms of \(x\), remember the substitution \(u = x^2 + 1 \implies x^2 = u-1\); hence \( x = \sqrt{u-1}\)
08

Perform the substitution

Substituting above expression of x into the integral and simplifying, we get: \(\int \frac{u-1}{u^2} (\frac{1}{2\sqrt{u-1}} du) = \frac{1}{2} \int \frac{\du}{u^2\sqrt{u-1}}\) Letting \(w^2=u-1\), we get \(\frac{1}{2}\int\frac{1}{(w^2+1)w^2}2wdx\). Finally this gives: \(\int \frac{1}{(w^2 + 1)w^2} dw\)
09

Calculating the integral and backward substitution

Now we compute the integral: Using partial fraction decomposition, \(\int\frac{1-\frac{w^2}{w^2+1}}{(w^2+1)w^2}dw=\int(1-\frac{1}{1+w^2})dw\) \(=w-\arctan(w)+C\) Now we substitute \(w\) back with \(\sqrt{u-1}\) and then substitute \(u\) back with \(x^2 + 1\): \(=\sqrt{x^2}-\arctan(\sqrt{x^2})+C\) \(=x-\arctan(x)+C\)
10

Verify the agreement of the results

After calculating the integral using two different substitutions, we have found the following results: Method (a): \(x-\arctan(x)+C\) Method (b): \(x-\arctan(x)+C\) As both results are the same, we have verified that the results agree.

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Three of these six antiderivatives are elementary. Find them. (A) \(\int x \cos x d x\) (B) \(\int \frac{\cos x}{x} d x\) (C) \(\int \frac{x d x}{\ln x}\) (D) \(\int \frac{\ln x^{2}}{x} d x\) (E) \(\int \sqrt{x-1} \sqrt{x} \sqrt{x+1} d x\) (F) \(\int \sqrt{x-1} \sqrt{x+1} x d x\)

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