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Question: If\(\int_{}^{}\frac{dx}{x^{3}(x - 1)^{1/2}}\)= \(\frac{\sqrt{x - 1}(3x + 2)}{4x^{2}}\) + K tan<sup>–...

If∫dxx3(x−1)1/2\int_{}^{}\frac{dx}{x^{3}(x - 1)^{1/2}}= x−1(3x+2)4x2\frac{\sqrt{x - 1}(3x + 2)}{4x^{2}} + K tan–1x−1\sqrt{x - 1} + c then the value of K is –

A

½

B

1

C

¼

D

¾

Answer

¾

Explanation

Solution

Put x – 1 = t2 so that dx = 2t dt, and

∫dxx3(x−1)1/2\int_{}^{}\frac{dx}{x^{3}(x - 1)^{1/2}}= ∫2tdt(t2+1)3t\int_{}^{}\frac{2tdt}{(t^{2} + 1)^{3}t}= 2∫dt(t2+1)3\int_{}^{}\frac{dt}{(t^{2} + 1)^{3}}

Now we formula of ∫dy(y2+k2)n\int_{}^{}\frac{dy}{(y^{2} + k^{2})^{n}} to write

∫dt(t2+1)3\int_{}^{}\frac{dt}{(t^{2} + 1)^{3}} = t4(t2+1)2\frac{t}{4(t^{2} + 1)^{2}} + 34\frac { 3 } { 4 } ∫dt(t2+1)2\int_{}^{}\frac{dt}{(t^{2} + 1)^{2}}and

∫dt(t2+1)2\int_{}^{}\frac{dt}{(t^{2} + 1)^{2}} = t2(t2+1)\frac{t}{2(t^{2} + 1)} +12\frac { 1 } { 2 } ∫dt(t2+1)\int_{}^{}\frac{dt}{(t^{2} + 1)}= t2(t2+1)\frac{t}{2(t^{2} + 1)}+12\frac{1}{2}tan–1 t

Hence ∫dxx3(x−1)1/2\int_{}^{}\frac{dx}{x^{3}(x - 1)^{1/2}}= t2(t2+1)2\frac{t}{2(t^{2} + 1)^{2}}+3t4(t2+1)\frac{3t}{4(t^{2} + 1)}+34\frac{3}{4} tan–1 t + c

= 14(t2+1)2\frac{1}{4(t^{2} + 1)^{2}} (2t + 3t3 + 3t) + 34\frac{3}{4} tan–1 t + c

=x−1(3x+2)4x2\frac{\sqrt{x - 1}(3x + 2)}{4x^{2}}+34\frac{3}{4}tan–1 x−1\sqrt{x - 1} + c