Whats wrong with this integral?

Question

$$\int \frac{(x+7)dx}{x^2+2x+5}$$

completing the square we get:

$$(x^2+2x+1)+5-1$$ $$(x+1)^2+4$$ then: $$U = x + 1$$ $$a = 2$$ hence:

$$\int \frac{(U+6)dU}{U^2+a^2}$$

using trig substitution $U = a\tan\theta$ and $dU = a\sec^2\theta d\theta$:

$$\int a\tan\theta d\theta + 6\int d\theta$$ $$a\ln(\sec\theta) + 6\theta$$ $$a\ln\left(\frac{\sqrt{a^2+U^2}}{a}\right) + 6\tan^{-1}\left(\frac{U}{a}\right)$$ $$2\ln\left(\frac{\sqrt{x^2+2x+5}}{2}\right)+ 6\tan^{-1}\left(\frac{x+1}{2}\right) $$

what did I get wrong?

book answer:

$$\frac{1}{2}\ln(x^2+2x+5)+ 3\tan^{-1}\left(\frac{x+1}{2}\right)$$

Answer 1

The mistake is here

$$\int \frac{(U+6)dU}{U^2+a^2}=\color{red}{\int a\tan\theta d\theta + 6\int d\theta}$$

we should have

$$\int \frac{(U+6)dU}{U^2+a^2}=\int \tan\theta d\theta + \frac 6a\int d\theta$$

since

$$\frac{dU}{U^2+a^2}=\frac{\frac a{\cos^2 \theta}d\theta}{\frac {a^2}{\cos^2 \theta}}=\frac1ad\theta$$

We obtain indeed

$$\ln\left(\frac{\sqrt{x^2+2x+5}}{2}\right)+ 3\tan^{-1}\left(\frac{x+1}{2}\right)$$

with

$$\ln\left(\frac{\sqrt{x^2+2x+5}}{2}\right)=\frac12\ln\left(\frac{x^2+2x+5}{4}\right)=\frac12\ln\left(x^2+2x+5\right)+C$$

Answer 2

Consider$$\int\frac{U+6}{U^2+4}\,\mathrm dU.$$If you do $U=2\tan\theta$ and $\mathrm dU=2(1+\tan^2\theta)\,\mathrm d\theta$, what you do get is$$\int\tan\theta\,\mathrm d\theta+\int3\,\mathrm d\theta.$$Besides,$$\int\tan\theta\,\mathrm d\theta=-\log(\cos\theta).$$

Answer 3

Another possible approach: \begin{align*} \int\frac{x + 7}{x^{2} + 2x + 5}\mathrm{d}x & = \frac{1}{2}\int\frac{2x + 14}{x^{2} + 2x + 5}\mathrm{d}x\\\\ & = \frac{1}{2}\int\frac{2x + 2}{x^{2} + 2x + 5}\mathrm{d}x + \frac{1}{2}\int\frac{12}{x^{2} + 2x + 5}\mathrm{d}x\\\\ & = \frac{1}{2}\int\frac{\mathrm{d}(x^{2} + 2x + 5)}{x^{2} + 2x + 5} + 3\int\frac{\mathrm{d}((x + 1)/2)}{((x + 1)/2)^{2} + 1}\\\\ & = \frac{1}{2}\ln(x^{2} + 2x + 5) + 3\arctan\left(\frac{x + 1}{2}\right) + C \end{align*}