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Question: \[\int_{}^{}{(2x - 5)\sqrt{x^{2} - 4x + 3}}dx =\]...

∫(2x−5)x2−4x+3dx=\int_{}^{}{(2x - 5)\sqrt{x^{2} - 4x + 3}}dx =

A

log⁡∣x−2+x2−4x+3∣+c\log|x - 2 + \sqrt{x^{2} - 4x + 3}| + c

B

log⁡∣x−2−x2−4x+3∣+c\log|x - 2 - \sqrt{x^{2} - 4x + 3}| + c

C

log⁡∣x+2+x2−4x+3∣+c\log|x + 2 + \sqrt{x^{2} - 4x + 3}| + c

D

None of these

Answer

None of these

Explanation

Solution

Let 2x−5=Mddx(x2−4x+3)+N⇒2x−5=M(2x−4)+N2x - 5 = M\frac{d}{dx}(x^{2} - 4x + 3) + N \Rightarrow 2x - 5 = M(2x - 4) + N

Equating the coefficients of x and constant terms on both sides, we get

2=2M⇒M=12 = 2M \Rightarrow M = 1

−5=−4M+N- 5 = - 4M + N ⇒N=4M−5=−1\Rightarrow N = 4M - 5 = - 1

Hence, I=∫{(2x−4)−1}x2−4x+3dxI = \int_{}^{}\left\{ (2x - 4) - 1 \right\}\sqrt{x^{2} - 4x + 3}dx

⇒I=∫(2x−4)x2−4x+3dx−∫x2−4x+3dx\Rightarrow I = \int_{}^{}{(2x - 4)\sqrt{x^{2} - 4x + 3}}dx - \int_{}^{}\sqrt{x^{2} - 4x + 3}dx I=I1−I2(say)I = I_{1} - I_{2}(say)

I1=∫(2x−4)x2−4x+3dxI_{1} = \int_{}^{}{(2x - 4)\sqrt{x^{2} - 4x + 3}}dx =23(x2−4x+3)3/2+c1= \frac{2}{3}(x^{2} - 4x + 3)^{3/2} + c_{1}

I2=∫x2−4x+3dx=∫x2−4x+4−4+3dxI_{2} = \int_{}^{}\sqrt{x^{2} - 4x + 3}dx = \int_{}^{}{\sqrt{x^{2} - 4x + 4 - 4 + 3}dx}

=∫(x−2)2−12⇒= \int_{}^{}\sqrt{(x - 2)^{2} - 1^{2}} \Rightarrow I2=12(x−2)x2−4x+3−12.12log⁡{(x−2)+x2−4x+3}I_{2} = \frac{1}{2}(x - 2)\sqrt{x^{2} - 4x + 3}\frac{- 1}{2}.1^{2}\log\left\{ (x - 2) + \sqrt{x^{2} - 4x + 3} \right\} =12(x−2)x2−4x+3= \frac{1}{2}(x - 2)\sqrt{x^{2} - 4x + 3} −12log⁡[(x−2)+x2−4x+3]+c2\frac{- 1}{2}\log\left\lbrack (x - 2) + \sqrt{x^{2} - 4x + 3} \right\rbrack + c_{2}, Therefore I=I1−I2I = I_{1} - I_{2}

I=23(x2−4x+3)3/2+12(−2)x2−4x+3−12log⁡[(x−2)+x2−4x+3]+cI = \frac{2}{3}(x^{2} - 4x + 3)^{3/2} + \frac{1}{2}( - 2)\sqrt{x^{2} - 4x + 3} - \frac{1}{2}\log\left\lbrack (x - 2) + \sqrt{x^{2} - 4x + 3} \right\rbrack + c (Where c=c1+c2)(\text{Where }c = c_{1} + c_{2})