I need to integrate the following function:
$$\int (4x^2 +3x+7+\, \frac5{3x} -\frac3{x^2})dx$$
I'm trying to do it term by term by using the exponent rule:
$$\int x^ndx = \frac1{n+1}x^{n+1}$$
So I get:
$$\int4x^2dx = \frac43x^3+C$$
$$\int3xdx = \frac32x^2+C$$ $$\int7dx = 7x + C$$
But I have no idea as how I should integrate the last two terms with divisors. Could I have some advise?
Treat $$-\frac{3}{x^2}$$As: $$-3\cdot x^{-2}$$So the $\frac{1}{n+1}x^{n+1}$ rule works fine: $$-\frac{3}{(-2)+1)}x^{(-2)+1}=3\cdot x^{-1}=\frac{3}{x}$$Is your desired integral. $1/x$ is a bit tricky as $n=-1$, so $\frac{1}{n+1}=\frac{1}{0}$. This rule doesn't work here, also because $\frac{d}{dx}x^0=0$ not $x^{-1}$. We have a new ""rule"" for this case: $$\int_1^x\frac{1}{t}\,\mathrm{d}t=\ln t$$Where appears the natural logarithm. You can treat this as a definition of $\ln$, or as a corollary of other definitions - your choice. Either way, it is the antiderivative of $1/x$ (for $x\gt0$ - if $x\lt 0$ you need to deal with $\ln|x|$)