We #160;have,I= #8747;2x+1 #160;dxx+2 #160;x 32Let #160;2x+1x+2 #160;x 32=Ax+2+Bx 3+Cx 32 #8658;2x+1x+2 #160;x 32=A #160;x 32+B #160;x+2 # ...

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Relation between AM,GM,HM for 2 Numbers

∫ 2 x+1 x+2 x...

Question

$\int \frac{2 x + 1}{(x + 2) {(x - 3)}^{2}} d x$

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Solution

$We have, I = \int \frac{(2 x + 1) d x}{(x + 2) {(x - 3)}^{2}} Let \frac{2 x + 1}{(x + 2) {(x - 3)}^{2}} = \frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{{(x - 3)}^{2}} \Rightarrow \frac{2 x + 1}{(x + 2) {(x - 3)}^{2}} = \frac{A {(x - 3)}^{2} + B (x + 2) (x - 3) + C (x + 2)}{(x + 2) {(x - 3)}^{2}} \Rightarrow 2 x + 1 = A (x^{2} - 6 x + 9) + B (x^{2} - x - 6) + C (x + 2) \Rightarrow 2 x + 1 = (A + B) x^{2} + (- 6 A - B + C) x + (9 A - 6 B + 2 C) Equating the coefficients of like terms A + B = 0 . . . . . (1) - 6 A - B + C = 2 . . . . . (2) 9 A - 6 B + 2 C = 1 . . . . . (3) Solving (1), (2) and (3), we get A = - \frac{3}{25}, B = \frac{3}{25} and C = \frac{7}{5} ∴ \frac{(2 x + 1) d x}{(x + 2) {(x - 3)}^{2}} = - \frac{3}{25 (x + 2)} + \frac{3}{25 (x - 3)} + \frac{7}{5 {(x - 3)}^{2}} \Rightarrow I = - \frac{3}{25} \int \frac{d x}{x + 2} + \frac{3}{25} \int \frac{d x}{x - 3} + \frac{7}{5} \int {(x - 3)}^{- 2} d x = - \frac{3}{25} \log |x + 2| + \frac{3}{25} \log |x - 3| + \frac{7}{5} [\frac{{(x - 3)}^{- 1}}{- 1}] + C = - \frac{3}{25} \log |x + 2| + \frac{3}{25} \log |x - 3| - \frac{7}{5 (x - 3)} + C$

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