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Properties of Inequalities

∫ 2 x+1 x+1 x...

Question

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

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Solution

$\int \frac{(2 x + 1)}{(x + 1) (x - 2)} d x Let \frac{2 x + 1}{(x + 1) (x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2} . . . . (1) \Rightarrow \frac{2 x + 1}{(x + 1) (x - 2)} = \frac{A (x - 2) + B (x + 1)}{(x + 1) (x - 2)} Then, (2 x + 1) = A (x - 2) + B (x + 1) . . . . (2) Putting (x - 2) = 0 or, x = 2 in eq (2) \Rightarrow 2 \times 2 + 1 = A \times 0 + B (2 + 1) \Rightarrow B = \frac{5}{3} Putting (x + 1) = 0 or, x = - 1 in eq (2) 2 \times - 1 + 1 + A (- 1 - 2) + B \times 0 \Rightarrow - 1 = A (- 3) \Rightarrow A = \frac{1}{3} Substituting the values of A and B in eq (1), we get ∴ \frac{2 x + 1}{(x + 1) (x - 2)} = \frac{1}{3} (x + 1) + \frac{5}{3} (x - 2) \int \frac{(2 x + 1) d x}{(x + 1) (x - 2)} = \frac{1}{3} \int \frac{1}{x + 1} d x + \frac{5}{3} \int \frac{1}{x - 2} d x = \frac{1}{3} \ln |x + 1| + \frac{5}{3} \ln |x - 2| + C$

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Choose the correct answer.
$\int \frac{1}{x^{2} + 2 x + 2} d x$ equals

$(a) x t a n^{- 1} (x + 1) + C (b) t a n^{- 1} (x + 1) + C (c) (x + 1) t a n^{- 1} x + C (d) t a n^{- 1} x + C$

Find : $\int \frac{(2 x - 5) e^{2 x}}{(2 x - 3)^{3}} d x .$ OR Find : $\int \frac{(x^{2} + x + 1)}{(x^{2} + 1) (x + 2)} d x .$

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