# Thomas' question at Yahoo! Answers regarding an indefinite integral

Staff member

#### MarkFL

Staff member
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.