#160; #8747;5xx+1 #160;x2 4dxLet5xx+1 #160;x 2 #160;x+2=Ax+1+Bx 2+Cx+2 #8658;5xx+1 #160;x 2 #160;x+2=A #160;x 2 #160;x+2+B #160;x+1 #16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Inequalities

∫ 5 x x+1 x 2...

Question

$\int \frac{5 x}{(x + 1) (x^{2} - 4)} d x$

Open in App

Solution

$\int \frac{5 x}{(x + 1) (x^{2} - 4)} d x Let \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{x + 2} \Rightarrow \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{A (x - 2) (x + 2) + B (x + 1) (x + 2) + C (x + 1) (x - 2)}{(x + 1) (x - 2) (x + 2)} \Rightarrow 5 x = A (x - 2) (x + 2) + B (x + 1) (x + 2) + C (x + 1) (x - 2) . . . . (1) Putting x - 2 = 0 or x = 2 in eq (1) \Rightarrow 5 \times 2 = B (2 + 1) (2 + 2) \Rightarrow B = \frac{10}{3 \times 4} = \frac{5}{6} Putting x + 2 = 0 or x = - 2 in eq (1) \Rightarrow 5 \times - 2 = C (- 2 + 1) (- 2 - 2) \Rightarrow \frac{- 10}{- 1 \times - 4} = C \Rightarrow C = \frac{- 5}{2} Putting x + 1 = 0 or x = - 1 in eq (1) \Rightarrow - 5 = A (- 1 - 2) (- 1 + 2) \Rightarrow \frac{- 5}{- 3} = A \Rightarrow A = \frac{5}{3} ∴ \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{5}{3} \times \frac{1}{x + 1} + \frac{5}{6 (x - 2)} - \frac{5}{2 (x + 2)} \Rightarrow \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{5}{6} \times \frac{2}{x + 1} + \frac{5}{6 (x - 2)} - \frac{5}{6} (\frac{3}{x + 2}) ∴ \int \frac{5 x}{(x + 1) (x - 2) (x + 2)} d x = \frac{5}{6} \int \frac{2}{x + 1} d x + \frac{5}{6} \int \frac{1}{x - 2} d x - \frac{5}{6} \int \frac{3}{x + 2} d x = \frac{5}{6} [2 \ln |x + 1| + \ln |x - 2| - 3 \ln |x + 2|] + C = \frac{5}{6} [\ln {|x + 1|}^{2} + \ln |x - 2| - \ln {|x + 2|}^{3}] + C = \frac{5}{6} \ln |\frac{{(x + 1)}^{2} (x - 2)}{{(x + 2)}^{3}}| + C$

Suggest Corrections