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Question: If \(\int_{}^{}\frac{dx}{x\sqrt{5x^{2} - 3}}\) = l (goƒ) (x) + C, then –...

If ∫dxx5x2−3\int_{}^{}\frac{dx}{x\sqrt{5x^{2} - 3}} = l (goƒ) (x) + C, then –

A

g(x) = tan–1x, ƒ(x) = 53x2−1\sqrt{\frac{5}{3}x^{2} - 1}, l = 13\frac{1}{\sqrt{3}}

B

g(x) = 53x2−1\sqrt{\frac{5}{3}x^{2} - 1}, ƒ(x) = tan–1 x, l = 13\frac{1}{\sqrt{3}}

C

g(x) = tan–1 x, ƒ(x) = 12\frac { 1 } { 2 } 5x2−3\sqrt{5x^{2} - 3}, l = 15\frac{1}{\sqrt{5}}

D

None of these

Answer

g(x) = tan–1x, ƒ(x) = 53x2−1\sqrt{\frac{5}{3}x^{2} - 1}, l = 13\frac{1}{\sqrt{3}}

Explanation

Solution

Put 5x2 – 3 = t2 Ž 5x dx = t dt

\ I = ∫xx25x2−3\int_{}^{}\frac{x}{x^{2}\sqrt{5x^{2} - 3}}dx = 15\frac { 1 } { 5 } ∫5tdt(t2+3)t\int_{}^{}\frac{5tdt}{(t^{2} + 3)t}

=

= 13\frac { 1 } { \sqrt { 3 } } tan–1 t3\frac{t}{\sqrt{3}} + C = 13\frac{1}{\sqrt{3}} tan–1 5x2−33\sqrt{\frac{5x^{2} - 3}{3}} + C.

\ g(x) = tan–1 x and ƒ(x) = 53x2−1\sqrt{\frac{5}{3}x^{2} - 1}