I am looking for a solution/explanation as to how to solve the following integral (sorry I am not familiar with the math language used here);
$$\int {1\over x^2+8x-3}\quad dx$$
I solved this problem a couple times and got the solution;
$$\frac{-19}{\sqrt{19}}\text{ln}\left|\frac{x+4+\sqrt{19}}{x^2+8x-3}\right|$$
I am fairly confident in this solution, however, symbolab and wolframalpha disagree. I am not sure if this is simply an alternate form of the solution or if it is incorrect altogether. I began solving the problem by completing the square and making the substitution $x+4=\sqrt{19}\sec(\theta)$ and $dx=\sqrt{19}\sec(\theta)\tan(\theta)$. From here I was able to pull out the constant $\frac{\sqrt{19}}{19}$, cancel out a $\tan(\theta)$ and was left with he constant times the integral of $\frac{\sec(\theta)}{\tan(\theta)}$. This simplifies to $\csc(\theta)$. I integrated this with the "multiply by $1/1$" method and used the substitution $u=\csc(\theta)+\cot(\theta)$ and $-du=\csc^2(\theta)+\csc(\theta)\cot(\theta)$. After solving, replacing my substitution, and using the trig identity, $x+4=\sqrt{19}\sec(\theta)$, I got my answer.
If anyone could help me determine if this answer is a correct form or how to solve it correctly it would be much appreciated. Again, sorry for the math formatting.
Best Answer
This quadratic has two real roots, $x = -4\pm\sqrt{19}$, so this is better handled with partial fractions: $$ \int \frac{dx}{x^2 + 8x -3} = \int\left[\frac{1}{x+4-\sqrt{19}}-\frac{1}{x+4+\sqrt{19}}\right]\frac{dx}{2\sqrt{19}} = \frac{1}{2\sqrt{19}}\ln\left|\frac{x+4-\sqrt{19}}{x+4+\sqrt{19}}\right| $$