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I'm going through the MIT online calc course, and I'm trying to find the following from an Integration Problem set:

$$\int \frac{x^2+1}{x^2+2x+2}dx$$

I could use the substitution u = x+1, which results in the correct answer:

$$\int \frac{(u-1)^2+1}{u^2+1}du$$ $$\int 1 + \frac{1}{u^2+1} - \frac{2u}{u^2+1}du$$ $$u + arctan \; u - ln(u^2+1)$$ $$x+1+arctan \; (x+1) - ln(x^2+2x+2)$$

But originally I tried using completing the square and the substitution $(tan \theta = x + 1)$. I get:

$$\int \frac{x^2+1}{(x+1)^2+1}dx$$ $$\int \frac{tan^2 \theta - 2tan \theta + 1 + 1}{sec^2\theta}sec^2\theta d\theta$$ $$\int sec^2\theta - 2tan\theta+1 d\theta$$ $$tan \theta + 2ln(cos \theta) + \theta$$ $$x + 1 - 2ln(x^2+2x+2) + arctan(x+1)$$

There's an extra 2 coefficient there. I don't think there's a math error, so I'm thinking the ambitious substitution is the cause. Any ideas?

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1 Answer 1

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The mistake is in thinking $\cos\theta=x^2+2x+2$.

Remark: I would probably want to divide first, getting $1-\dfrac{2x+1}{(x+1)^2+1}$.

Then write $\dfrac{2x+1}{(x+1)^2+1}$ as $\dfrac{2x+2}{(x+1)^2+1}-\dfrac{1}{(x+1)^2+1}$.

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  • $\begingroup$ Gah. GAH. Thank you! $\endgroup$ Commented Aug 31, 2013 at 0:58
  • $\begingroup$ You are welcome. $\endgroup$ Commented Aug 31, 2013 at 1:02
  • $\begingroup$ Why is there a plus sign in your last statement? 2x+1 is not 2x+2 + 1 ?? $\endgroup$ Commented Aug 31, 2013 at 1:42
  • $\begingroup$ @imranfat: Because I have trouble with minus signs. Thanks! $\endgroup$ Commented Aug 31, 2013 at 1:50

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