Solve the intergal....... (2x-1)/[x(x-2)(x-3)]........ Please help fast?

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Answer 1 · 2018-06-21T17:45:28

$\frac{5}{3} ln | (x - 3) | - \frac{3}{2} ln | (x - 2) | - \frac{1}{6} ln | x | + C, or,$ $5/3ln|(x-3)|-3/2ln|(x-2)|-1/6ln|x|+C, or,$

$ln ∣ ∣ ∣ ∣ \frac{{(x - 3)}^{\frac{5}{3}}}{{(x - 2)}^{\frac{3}{2}} x^{\frac{1}{6}}} ∣ ∣ ∣ ∣ + C$ $ln|(x-3)^(5/3)/{(x-2)^(3/2)x^(1/6)}|+C$ .

Suppose that, $I = \int \frac{2 x - 1}{x (x - 2) (x - 3)} d x$ $I=int(2x-1)/{x(x-2)(x-3)}dx$ .

$:. I=int{(2x)/{x(x-2)(x-3)}+(-1)/{x(x-2)(x-3)}}dx$ ,

$=int{2/{(x-2)(x-3)}+{(x-3)-(x-2)}/{x(x-2)(x-3)}}dx$ ,

$=int{2/{(x-2)(x-3)}+(x-3)/{x(x-2)(x-3)}-(x-2)/{x(x-2)(x-3)}}dx$ ,

$=2int1/{(x-2)(x-3)}dx+int1/{x(x-2)}dx-1/{x(x-3)}dx$ ,

$=2int{(x-2)-(x-3)}/{(x-2)(x-3)}dx+1/2int{x-(x-2)}/{x(x-2)}dx$

$-1/3int{x-(x-3)}/{x(x-3)}dx$ ,

$=2int{1/(x-3)-1/(x-2)}dx+1/2int{1/(x-2)-1/x}dx$

$-1/3int{1/(x-3)-1/x}dx$ ,

$=2ln|(x-3)|-2ln|(x-2)|+1/2ln|(x-2)|-1/2ln|x|-1/3ln|(x-3)|+1/3ln|x|$ ,

$=(2-1/3)ln|(x-3)|-(2-1/2)ln|(x-2)|-(1/2-1/3)ln|x|$ .

$rArr I=5/3ln|(x-3)|-3/2ln|(x-2)|-1/6ln|x|+C, or,$

$I=ln|(x-3)^(5/3)/{(x-2)^(3/2)x^(1/6)}|+C$ .

Answer 2 · 2018-06-21T17:48:22

$5/3ln|x-3|-3/2*ln|x-2|-1/6*ln|x|+C$

Converting your Integrand in partial fractions we get

$(2x-1)/(x*(x-2)(x-3))=5/(3*(x-3))-3/(2*(x-2))-1/(6*x)$

The result is to be seen above.