# Thomas' question at Yahoo Answers regarding an indefinite integral

• MHB
• MarkFL
In summary, the conversation is about a calculus question regarding the integration of arctan(x^1/2). The response includes a step-by-step explanation and solution of the problem using substitution and integration by parts. The responder also invites others to post more calculus problems in the forum.
MarkFL
Gold Member
MHB
Here is the question:

Calc 2 integral question?

Integrate: arctan(x^1/2) dx
Thanks!

Here is a link to the question:

Calc 2 integral question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

Hello Thomas,

We are give to evaluate:

$$\displaystyle I=\int\tan^{-1}\left(\sqrt{x} \right)\,dx$$

I would begin by rewriting the integral in preparation for a substitution:

$$\displaystyle I=2\int\sqrt{x}\tan^{-1}\left(\sqrt{x} \right)\,\frac{1}{2\sqrt{x}}dx$$

Now, use the substitution:

$$\displaystyle w=\sqrt{x}\,\therefore\,dw=\frac{1}{2\sqrt{x}}dx$$

and we have:

$$\displaystyle I=2\int w\tan^{-1}(w)\,dw$$

Now, using integration by parts, let:

$$\displaystyle u=\tan^{-1}(w)\,\therefore\,du=\frac{1}{w^2+1}\,dw$$

$$\displaystyle dv=w\,dw\,\therefore\,v=\frac{1}{2}w^2$$

and we have:

$$\displaystyle I=2\left(\frac{1}{2}w^2\tan^{-1}(w)-\frac{1}{2}\int\frac{w^2}{w^2+1}\,dw \right)$$

Distribute the 2, and rewrite the numerator of the integrand:

$$\displaystyle I=w^2\tan^{-1}(w)-\int\frac{(w^2+1)-1}{w^2+1}\,dw$$

$$\displaystyle I=w^2\tan^{-1}(w)-\int1-\frac{1}{w^2+1}\,dw$$

Now complete finding the anti-derivative:

$$\displaystyle I=w^2\tan^{-1}(w)-w+\tan^{-1}(w)+C$$

Factor for simplicity of expression:

$$\displaystyle I=(w^2+1)\tan^{-1}(w)-w+C$$

Back-substitute for $w$:

$$\displaystyle I=(x+1)\tan^{-1}\left(\sqrt{x} \right)-\sqrt{x}+C$$

To Thomas and any other guests viewing this topic, I invite and encourage you to post other calculus problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

## 1. What is an indefinite integral?

An indefinite integral is the process of finding the general antiderivative of a given function. It is denoted by the symbol ∫ and represents the area under a curve.

## 2. How is an indefinite integral different from a definite integral?

A definite integral has specific limits of integration, while an indefinite integral does not. This means that a definite integral will give a specific numerical value, while an indefinite integral will give a general formula.

## 3. What is the process for solving an indefinite integral?

The process for solving an indefinite integral involves using integration techniques, such as substitution or integration by parts, to find the antiderivative of the given function. This will result in a general formula with a constant of integration.

## 4. Can an indefinite integral have multiple solutions?

Yes, an indefinite integral can have multiple solutions because the constant of integration can take on any value. This means that there can be an infinite number of antiderivatives for a given function.

## 5. How is an indefinite integral used in real life?

An indefinite integral has many practical applications, such as finding the displacement, velocity, or acceleration of an object using the fundamental theorem of calculus. It is also used in various fields of science, including physics, engineering, and economics.

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