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Integration by Partial Fractions

∫√x-1/2x-3dx=

Question

$\int \sqrt{\frac{x - 1}{2 x - 3}} d x =$

A

s i n^{- 1} x + \sqrt{1 - x^{2}} + c

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B

s i n^{- 1} x - \sqrt{1 - x^{2}} + c

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C

2 s i n^{- 1} x - \sqrt{1 - x^{2}} + c

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D

\frac{1}{2} \sqrt{2 x^{2} - 5 x + 3} + \frac{1}{4} {cosh}^{- 1} (4 x - 5) + c

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Solution

The correct option is D $\frac{1}{2} \sqrt{2 x^{2} - 5 x + 3} + \frac{1}{4} {cosh}^{- 1} (4 x - 5) + c$

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Similar questions

\frac{x^{2} + 5 x + 1}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1) (x + 2)} + \frac{C}{(x + 1) (x + 2) (x + 3)}

\int {sin}^{- 1} x {cos}^{- 1} x d x = f^{- 1} (x) [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

.

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

\int {sin}^{- 1} x {cos}^{- 1} x d x = f^{- 1} (x) [A x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] + \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + c,

\frac{x^{2} + 5 x + 1}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1) (x + 2)}

+ \frac{C}{(x + 1) (x + 2) (x + 3)},

B

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