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Thomas' question at Yahoo! Answers regarding an indefinite integral

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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 Calculus forum.

Best Regards,

Mark.