How to solve $\int\frac{dx}{(x^2 - 8x + 1)}$?

Question

How to solve $\int\frac{dx}{(x^2 - 8x + 1)}$?

I can't spread it into partial fractions. I can't use u-substitution. And you can't do integration by parts. So, I'm wondering how to do it. Thanks for the help.

Answer 1

You can use partial fractions. It isn't necessary that the constants be integers for it to work. $$\frac{1}{x^2-8x+1} = \frac{1}{(x-4-\sqrt{15})(x-4+\sqrt{15})}$$ or if you prefer to substitute $u = x-4$ first: $$\frac{1}{u^2-15} = \frac{A}{u-\sqrt{15}} + \frac{B}{u+\sqrt{15}}$$

Answer 2

When there are no obvious roots, you can complete the square to arrive either

• to $\displaystyle\int \dfrac {du}{1+u^2}=\arctan(u)$
• or to $\displaystyle\int \dfrac{du}{1-u^2}=\operatorname{argth}(u)$

In the present case $x^2-8x+1=(x-4)^2-15$, so set $u=\dfrac{x-4}{\sqrt{15}}$ and check you get the second form above.

Answer 3

$\int\frac{dx}{x^2-8x+1}=\int\frac{dx}{(x-4)^2-15}$

$\int\frac{dx}{(x-4)^2-15}=\int\frac{dx}{(x-4-\sqrt{15})(x-4+\sqrt{15})}$

$\int\frac{dx}{(x-4-\sqrt{15})(x-4+\sqrt{15})}=\frac12(\int\frac{dx}{x-4-\sqrt{15}}+\int\frac{dx}{x-4+\sqrt{15}})$