How do you integrate $\int \frac{x}{x - 6} d x$ $int x/(x-6) dx$ using partial fractions?

Question

How do I find the partial fraction decomposition of $\frac{2 x}{(x + 3) (3 x + 1)}$ $(2x)/((x+3)(3x+1))$ ?
How do I find the partial fraction decomposition of $\frac{1}{x^{3} + 2 x^{2} + x}$ $(1)/(x^3+2x^2+x$ ?
How do I find the partial fraction decomposition of $\frac{x^{4} + 1}{x^{5} + 4 x^{3}}$ $(x^4+1)/(x^5+4x^3)$ ?
How do I find the partial fraction decomposition of $\frac{x^{4}}{x^{4} - 1}$ $(x^4)/(x^4-1)$ ?
How do I find the partial fraction decomposition of $\frac{t^{4} + t^{2} + 1}{(t^{2} + 1) {(t^{2} + 4)}^{2}}$ $(t^4+t^2+1)/((t^2+1)(t^2+4)^2)$ ?
How do I find the integral $\int \frac{t^{2}}{t + 4} d t$ $intt^2/(t+4)dt$ ?
How do I find the integral $\int \frac{x - 9}{(x + 5) (x - 2)} d x$ $int(x-9)/((x+5)(x-2))dx$ ?
How do I find the integral $\int \frac{1}{(w - 4) (w + 1)} d w$ $int1/((w-4)(w+1))dw$ ?
How do I find the integral $\int \frac{d x}{x^{2} {(x - 1)}^{2}}$ $intdx/(x^2(x-1)^2)$ ?
How do I find the integral $\int \frac{x^{3} + 4}{x^{2} + 4} d x$ $int(x^3+4)/(x^2+4)dx$ ?

6947 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-08T02:01:35

Don't need to use partial fractions.

Step 1: divide $(x - 6)$ $(x-6)$ into $x$ $x$ using long division:

$\frac{x}{x - 6} = 1 + \frac{6}{x - 6}$ $x/(x-6)=1+6/(x-6)$

Step 2: Integrate

$\int [1 + \frac{6}{x - 6}] d x = x + 6 ln (x - 6) + C$ $int[1+6/(x-6)]dx=x+6ln(x-6)+C$

hope that helped

How do you integrate ∫xx−6dxint x/(x-6) dx using partial fractions?