- #1
karush
Gold Member
MHB
- 3,269
- 5
partial fractions
$$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}
=\frac{A}{(x^2+5)}+\frac{B}{(x-3)}$$
$$3x^2+x+12=A(x-3)+B(x^2+5)$$
x=3 then 27+3+12=14B
3=B
x=0 then
12=-3A+15
1=A
$$\int\frac{1}{(x^2+5)} \, dx
+3\int\frac{1}{(x-3)}\, dx$$ $\displaystyle
\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}}
+3\ln\left(\left|x-3\right|\right)+C$
maybe? not sure
$$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}
=\frac{A}{(x^2+5)}+\frac{B}{(x-3)}$$
$$3x^2+x+12=A(x-3)+B(x^2+5)$$
x=3 then 27+3+12=14B
3=B
x=0 then
12=-3A+15
1=A
$$\int\frac{1}{(x^2+5)} \, dx
+3\int\frac{1}{(x-3)}\, dx$$ $\displaystyle
\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}}
+3\ln\left(\left|x-3\right|\right)+C$
maybe? not sure
Last edited: