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I need to complete the square on the following integral. Once this is done apparently I will be able to use on of the integration tables in the back of my book.

$\int x \sqrt{x^2 + 6x +3} dx $

This is what I have come up with so far:

$\int x \sqrt{(x+3)^2 -6} $

I really am at a loss.

Any help with this would be appreciated.

Thank you

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3 Answers

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Substitute $u=x+3 \implies x=u-3, dx=du$:$$\int x\sqrt{(x+3)^2-6}dx=\int (u-3)\sqrt{u^2-6} du$$Integrating the integral gives: $$\int u\sqrt{u^2-6}du-\int 3\sqrt{u^2-6} du=\frac{(u^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{u\sqrt{u^2-6}}{2}-3\ln\left(u+\sqrt{u^2-6}\right)\right)$$In terms of $x$:$$\int x\sqrt{(x+3)^2-6}dx=$$$$\frac{((x+3)^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{(x+3)\sqrt{(x+3)^2-6}}{2}-3\ln\left(x+3+\sqrt{(x+3)^2-6}\right)\right)$$

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$$ I= \int x \sqrt {x^2 +6x +3} dx $$ This is best done by substitution. $$I= \int x \sqrt {(x+3)^2-6} dx \\ x+3= \sqrt 6 \sec \theta \implies dx=\sqrt 6 \sec \theta \tan \theta d \theta$$ $$ \therefore I= \sqrt 6 \int \sec \theta \tan \theta ( \sqrt 6 \sec \theta -3) \sqrt {6( \sec^2 \theta -1)} \\ I=6 \int \sec \theta \tan^2 \theta ( \sqrt 6 \sec \theta -3)d \theta \\ I=6 \sqrt6 \int \sec^2 \theta \tan^2 \theta d\theta-18\int \sec \theta \tan^2 \theta d \theta \\ \therefore I= 2 \sqrt6 \tan^3 \theta - 18\int \sec^3 \theta d \theta + 18\int \sec \theta d \theta \\ I =2 \sqrt6 \tan^3 \theta -9 \sec \theta \tan \theta +9 \ln | \sec \theta + \tan \theta | +C$$

Unless I screwed up the algebra I think this is correct. It could also be done using the substitution $ u=x^2+6x+3 $ and evaluating $\int \sqrt {x^2+6x+3} dx$.

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Method using only completing the square:

Set up for substitution: $$ I= \int x \sqrt {(x+3)^2-6} dx= \int (x+3)\sqrt {(x+3)^2-6}dx -3\int\sqrt {(x+3)^2-6}dx$$

Substitute: $$u^2=(x+3)^2-6 \implies udu=(x+3)dx \\ \sqrt 6 v=x+3 \implies \sqrt 6 dv=dx$$

$$ \therefore I=\int u^2 du -18 \int \sqrt {v^2-1}dv $$

These are both elementary integrals.

This integral can also be done by parts... so many different methods.

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