- #1

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[tex]

\int \frac{(du)}{(a^2 + u^2)} = \frac{1}{a} \cdot arctan \frac{u}{a} + C

[/tex]

I was given the problem of solving the indefinite integral of

[tex]

\int \frac{(dx)}{(2x^2 + 2x + 5)}

[/tex]

First, I multiplied the integral by (1/2) / (1/2) to eliminate the coefficient of the x^2 in the denominator, so now I am left with

[tex]

\frac{1}{2} \cdot \int \frac{(dx)}{(x^2 + x + \frac{5}{2})}

[/tex]

Now, in completing the square of the denominator, I added [tex] \frac{1}{4} - \frac{1}{4} [/tex] (which is zero) so that the equation would look like this:

[tex]

= \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{5}{2} + \frac{1}{4} - \frac{1}{4}}

[/tex]

Simplifying, I got:

[tex]

= \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{1}{4} + (\frac{5}{2} - \frac{1}{4})}

[/tex]

[tex]

= \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + \frac{9}{4}}

[/tex]

[tex]

= \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + ( \frac{3}{2})^2}

[/tex]

If we let [tex](x + \frac{1}{2})^2 = u^2[/tex] and [tex]( \frac{3}{2})^2 = a^2[/tex] we now have the integrable form stated above, so

[tex]

= \frac{1}{2} \int \frac{du}{u^2 + a^2}

[/tex]

[tex]

= \frac{1}{2} \cdot \frac{2}{3} \cdot arctan \frac{u}{a} + C

[/tex]

[tex]

= \frac{1}{3} \cdot arctan \frac{x^2 + \frac{1}{2}}{\frac{3}{2}} + C

[/tex]

[tex]

= \frac{1}{3} \cdot arctan \frac{2(x^2 + \frac{1}{2})}{3} + C

[/tex]

My questions are:

(1) Is this the correct solution?

(2) If my solution is correct, how do you get the necessary constant in order to make the denominator of this problem a complete square? the (1/4 - 1/4) I added to the denominator just popped out of my mind. Is there any way to get it without resorting to trial and error?

thank you