Aid in identifying an error made whilst evaluating an integral

Question

Evaluate $$I=\int\frac{1}{x^2-8x+8}~dx$$

First, complete the square using the denominator: $$x^2-8x+8=(x-4)^2-8$$ and therefore, let $x=2\sqrt{2}\sec{\theta}+4$, $\therefore dx=2\sqrt{2}\sec{\theta}\tan{\theta}~d\theta$, and hence, we have $$I=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{(2\sqrt{2}\sec{\theta}+4-4)^2-8}~d\theta$$ $$=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{(2\sqrt{2}\sec{\theta})^2-8}~d\theta$$ $$=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{8\sec^2{\theta}-8}~d\theta$$ $$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}\tan{\theta}}{\sec^2\theta-1}~d\theta$$ $$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}\tan{\theta}}{\tan^2\theta}~d\theta$$ $$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}}{\tan\theta}~d\theta$$ note:$$\frac{\sec{\theta}}{\tan{\theta}}=\frac{\cos{\theta}}{\sin{\theta}}\cdot\frac{1}{\cos{\theta}}=\csc{\theta}$$ therefore, $$I=-\frac{\sqrt{2}}{4}\ln{|\csc\theta+\cot\theta|}+C$$ now, using the fact that $x=2\sqrt{2}\sec\theta+4$, we have the following identities, based on the definition of the trigonometric functions: $$\csc\theta=\frac{x-4}{\sqrt{x^2-8x+8}}$$ as well as $$\cot\theta=\frac{2\sqrt{2}}{\sqrt{x^2-8x+8}}$$ therefore, the final integral is given by: $$I=-\frac{\sqrt{2}}{4}\ln{\left|\frac{x+2\sqrt{2}-4}{\sqrt{x^2-8x+8}}\right|}+C$$ which, I've been informed, is incorrect. Would anyone be kind enough to help me realize my error? Any responses are appreciated. Thank you.

Answer 1

First of all consider the derivations of the inverse hyperbolic functions as follows

$$(\operatorname{artanh}(x))'=\frac{1}{1-x^2}~\text{for}~|x|<1~~~~(\operatorname{arcoth}(x))'=\frac{-1}{x^2-1}~\text{for}~|x|>1$$

These can be found by using the identity $x'(y)=\frac1{y'(x)}$ and some knowledge concerning the hyperbolic functions. Integrating these equations leads to some nice formulas for integrating fractions of the form $x^2-1$ and $1-x^2$. With these tools one can avoid using a trigonometric subsitution here.

Especially for your integral we get

$$\begin{align} \int\frac1{x^2-8x+8}dx&=\int\frac1{(x-4)^2-8}dx\\ &=\frac18\int\frac1{\left(\frac{x-4}{2\sqrt{2}}\right)^2-1}dx\\ &=\frac18\int\frac1{u^2-1}du\\ &=-\frac18\operatorname{artanh}(u)+C\\ &=-\frac18\operatorname{artanh}\left(\frac{x-4}{2\sqrt{2}}\right)+C \end{align}$$

The only thing you have to care about is when to use the inverse hyperbolic tangent and when to use the inverse hyperbolic cotangent.

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By using the definitions of the hyperbolic functions

$$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}~~~\coth(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$

we can verify the following formulas for the inverse functions

$$\operatorname{artanh}(x)=\frac12\ln\frac{1+x}{1-x}~\text{for}~|x|<1~~~\operatorname{arcoth}(x)=\frac12\ln\frac{x+1}{x-1}~\text{for}~|x|>1$$

which can be used to reshape the solution of the integral in terms of the natural logarthim.

Answer 2

Your answer is correct and it's only a matter of rewriting.

Let us consider your integral: $$I=\int\frac{1}{x^2-8x+8}\ dx.$$ I approached this problem by factoring the denominator instead of completing the square: $$x^2-8x+8=0.$$ The discriminant is given by $\Delta_x=(-8)^2-4(1)(8)=64-32=32$ such that $\sqrt{\Delta_x}=\sqrt{32}=4\sqrt{2}$.

Solutions for $x$ are given by $$x_{1,2}=\frac{-(-8)\pm4\sqrt{2}}{2(1)}=4\pm2\sqrt{2}.$$ Thus, $$x^2-8x+8=(x-4-2\sqrt{2})(x-4+2\sqrt{2}).$$ Next I applied partial fraction decomposition on the integrand: $$\frac{1}{x^2-8x+8}=\frac{A}{x-4-2\sqrt{2}}+\frac{B}{x-4+2\sqrt{2}}.$$ It follows that $$1=A(x-4+2\sqrt{2})+B(x-4-2\sqrt{2})=(A+B)(x-4)+(A-B)2\sqrt{2}.$$ It can be derived that $A+B=0$ and $(A-B)2\sqrt{2}=1$ which means that $A=\frac{1}{4\sqrt{2}}$ and $B=-\frac{1}{4\sqrt{2}}$.

Thus, $$\frac{1}{x^2-8x+8}=\frac{1}{4\sqrt{2}}\left(\frac{1}{x-4-2\sqrt{2}}-\frac{1}{x-4+2\sqrt{2}}\right).$$ Substitution into your integral gives $$I=\frac{1}{4\sqrt{2}}\int\left(\frac{1}{x-4-2\sqrt{2}}-\frac{1}{x-4+2\sqrt{2}}\right) dx.$$ Integration yields $$I=\frac{1}{4\sqrt{2}}\ln\left|\frac{x-4-2\sqrt{2}}{x-4+2\sqrt{2}}\right|+\text{constant}.$$ Okay, so we have arrived at an expression. Let us rewrite it to yield your expression.

Consider
$$I=-\frac{1}{4\sqrt{2}}\ln\left|\left(\frac{x-4-2\sqrt{2}}{x-4+2\sqrt{2}}\right)^{-1}\right|+\text{constant}.$$ It follows that $$I=-\frac{1}{4\sqrt{2}}\ln\left|\frac{x-4+2\sqrt{2}}{x-4-2\sqrt{2}}\right|+\text{constant}.$$ Within the argument of the natural logarithm; multiply both the denominator and numerator by $x-4+2\sqrt{2}$: $$I=-\frac{1}{4\sqrt{2}}\ln\left|\frac{x-4+2\sqrt{2}}{x-4-2\sqrt{2}}\frac{x-4+2\sqrt{2}}{x-4+2\sqrt{2}}\right|+\text{constant}.$$ It follows that $$I=-\frac{1}{4\sqrt{2}}\ln\left|\frac{(x-4+2\sqrt{2})^2}{x^2-8x+8}\right|+\text{constant}.$$
Write $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$: $$I=-\frac{\sqrt{2}}{4}\frac{1}{2}\ln\left|\frac{(x-4+2\sqrt{2})^2}{x^2-8x+8}\right|+\text{constant}.$$ It follows that $$I=-\frac{\sqrt{2}}{4}\ln\left|\frac{x-4+2\sqrt{2}}{\sqrt{x^2-8x+8}}\right|+\text{constant}.$$ We have obtained your expression.