Question

Calculus problem:

Answer 1

Split the integrand into partial fractions.

$\dfrac{9x+2}{x^2+x-6} = \dfrac{9x+2}{(x-2)(x+3)} = \dfrac a{x-2} + \dfrac b{x+3}$

$\implies 9x+2 = a(x+3) + b(x-2) = (a+b)x + (3a-2b)$

$\implies \begin{cases}a+b=9 \\ 3a-2b=2\end{cases} \implies a=4,b=5$

Then we have

$\displaystyle \int \frac{9x+2}{x^2+x-6} \, dx = 4 \int \frac{dx}{x-2} + 5 \int \frac{dx}{x+3} \\\\ = \boxed{4\ln|x-2| + 5\ln|x+3| + C}$

which follows from the result

$\displaystyle \int \frac{dx}x = \ln|x|+C$

Answer 2

Don’t think it’s possible