Question

Calculus problem:


`\int\limits_ } \, (9x+2)/(x^2+x-6) dx`
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Answer 1

Don’t think it’s possible

Answer 2

Split the integrand into partial fractions.

`(9x+2)/(x^2+x-6) = (9x+2)/((x-2)(x+3)) = \frac a{x-2} + \frac 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 (9x+2)/(x^2+x-6) \, dx = 4 \int (dx)/(x-2) + 5 \int (dx)/(x+3) \\\\ = \boxedx+3`

which follows from the result

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