Find the integral?

Question

Evaluate the integral 7/(w^2+3w+3) dw.

Here's my work:

I've completed the square and came up with this:

∫ 7/[(w+3/2)^2+3/4] dw

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where u=w+3/2 and du=dw.

So ∫ 7/(u^2+3/4) du

and now I got stucked. What do I do from here? How do I get to the answer? I know that I'm supposed to use the inverse tangent, since ∫ dx/(1+x^2)=arctanx+C.

Answer 1

∫ [ du / (u² + a² ] = (1/a)Arctan(u / a) + C

Let u = w + 3/2. Then du = dw

So, 7∫ [ du / (u² + (√3/2)²)] = (14/√3) Arctan [ (2u / √3) ] + C

= (14/√3) Arctan [ (2w +3) / √3 ] + C

Answer 2

You are on the right track.

Just make the substitution u=w+(3/4), du=dw and the integral becomes

7∫du/(u²+3/4) which integrates directly into the arctan--the 3/4 just makes it messy!