Integral question - $\int\frac{(x+6)\,dx}{4x-x^2}$
What I did is $$\int\frac{(x+6)\,dx}{x(4-x)}$$
then
$$\int\frac{(x+6)\,dx}{4x-x^2}= \int\left(\frac{A \,}{x}+\frac{B}{4-x}\right) dx$$
this is the right way?
Thanks!
Addition:
$$A(4-x)+B(x)=x+6$$
$$ x=0 , 4A=6 => A=\frac{6}{4} , x=4, 4B=4+6 => B=\frac{10}{4}$$
$$\int\left(\frac{6\,}{4x}+\frac{10}{4(4-x)} \right)dx$$
Yes, you've done a great job, having just learned the partial-fraction technique!
I suspect you can take the integrals from here, and evaluate?
First, you can also simplify the constant terms (divide numerator/denominator by $2$ in each integral), or putting $A = \dfrac 32, \; B = \dfrac 52$, and you remove them from inside the integral (multiply each integral by its respective constant term.)
