Answer:
[tex]\int\limits {\frac{x}{x^2-8x+32}} \, dx=\frac{1}{2}ln|(x-4)^2+16|+4tan^{-1}(\frac{1}{16}(x-4))[/tex]
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Step-by-step explanation:
lol. why is this problem so hard.
but basically, my first instinct when seeing a fraction was to use partial fraction, but that failed.
So next option was to complete the square. After completing the square, I used u-substitution and seperated my integral into 2 different fractions. Both of which requires you to do double substitution.
Hopefully you already know the integral for ln|x| and tan⁻¹(x) which are used in solving this problem.
