How do I take the intergral of (8x+11)/(x^2+9) dx?

Question

How do I take the intergral of (8x+11)/(x^2+9) dx?

Answer 1

I = 4 ∫ ( 2x + 11/4 ) / (x² + 9) dx
I = 4 ∫ (2x) / (x² + 9) dx + 11 ∫ 1 / (x² + 3²) dx
I = 4 log (x² + 9) + (11/3) tan^(-1) (x / 3) + C

Answer 2

Split into 2 fractions:
∫ (8x+11)/(x²+9) dx
∫ (8x)/(x²+9) dx + ∫ 11/(x²+9) dx
8∫ x/(x²+9) dx + 11∫ 1/(x²+9) dx

Use u-substitution for the first integral.
u = x²+9
du = 2x dx
du/2 = x dx
8∫ 1/(u) du/2
4 ln|u| = 4ln|x²+9|

The second integral is in the form ∫1/(u²+a²) dx which is
1/a * arctan(u/a).
So the second integral is 11* 1/3 * arctan(x/3)

So the final answer is:
4ln|x²+9| + 11/3 arctan(x/3) + C

Answer 3

∫8x^3+11x^2+72x +99 dx= 2x^4+11/3 x^3 +36x^2 +99x + c

Answer 4

Split it into
8x/(x^2+9) + 11/(x^2+9)

The first one is ln(), and the second one is arctan()