IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

ashish_banga

if $\int x log( x + \frac{1}{x}) dx = F(x) log ( x + 1) + G(x) {x}^{2} + L x + C$

then find F(x), G(x), L and C

aditi_g

just apply ILATE taking log(x+1/x) as 1st function and x as second and ull get

F(x)= (x^2-1)/2

g(x)= -1/2 logx

L=1/2

lemme know if u have any prob wid this ans ill post the whole solution

aditi_g

C is a constant how can we determine its value??
i dont think the value of C shud be asked

ashish_banga

i dont know the answer correctly it is very faded

allamraju

Using Integration by parts,we get

(x²/2)ln(x+1/x)- $\int$ x(x²-1)/2(x²+1)dx

='' '' -1/4 $\int$ (t-1/t+1)dt where t=x²

=" " -1/4 $\int$ (1-2/t+1)dt

=(x²/2)ln(x²+1)-x²/2lnx-x²/4+1/2ln(x²+1)+C(since t=x²)

=(x²+1)/2ln(x²+1)-x²/4(2lnx+1)+C

First of all,tell me whether my integration is correct or is there any mistake.Also kindly check your question once again regarding the thing after the equality in your question.

Blazzer

I think there is a mistake in the question . The correct question shud be :

$\int[xlog(1+\frac{1}{x})]dx$

Considering the above case ,

Proceeding , we have ,

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

Hence $log(1+x)(\frac{{x}^{2}}{2})-(logx)(\frac{{x}^{2}}{2})+\frac{1}{2}\int(\frac{x+1-1}{x+1})$

$log(1+x)(\frac{{x}^{2}}{2})-(logx)(\frac{{x}^{2}}{2})+\frac{1}{2}x-\frac{1}{2}log(x+1)$

Hence finally we get ,

$(\frac{{x^{2}-1}}{2})log(1+x)-(logx)(\frac{{x}^{2}}{2})+\frac{1}{2}x$

Hence we get ,

$F(x)=\frac{{x}^{2}-1}{2},\;$ $G(x) = \frac{-(logx)}{2}$ $and\;L=\frac{1}{2}$

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Using Integration by parts,we get

(x2/2)ln(x+1/x)-x(x2-1)/2(x2+1)dx

='' '' -1/4 (t-1/t+1)dt where t=x2

=" " -1/4(1-2/t+1)dt