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Mathematics Question on Methods of Integration

If = ∫xlog⁡(1+1x)dx=f(x)log⁡(x+1)+g(x)x2+Lx+C\int x \log\left(1+ \frac{1}{x}\right)dx = f\left(x\right)\log\left(x+1\right)+g\left(x\right)x^{2}+Lx +C, then

A

f(x)=12x2f(x) = \frac{1}{2} x^2

B

g(x)=log⁡xg(x) = \log x

C

L=1L = 1

D

None of these

Answer

None of these

Explanation

Solution

∫xlog⁡(1+1x)dx\int x \log \left(1+\frac{1}{x}\right) d x
=∫xlog⁡(x+1)dx−∫xlog⁡xdx=\int x \log (x+1) d x-\int x \log x d x
=x22log⁡(x+1)−12∫x2x+1dx−x22log⁡x+12∫x2xdx=\frac{x^{2}}{2} \log (x+1)-\frac{1}{2} \int \frac{x^{2}}{x+1} d x-\frac{x^{2}}{2} \log x+\frac{1}{2} \int \frac{x^{2}}{x} d x
=x22log⁡(x+1)−12∫(x−1+1x+1)dx−x22log⁡x+14x2=\frac{x^{2}}{2} \log (x+1)-\frac{1}{2} \int\left(x-1+\frac{1}{x+1}\right) d x-\frac{x^{2}}{2} \log x+\frac{1}{4} x^{2}
=x22log⁡(x+1)−x22log⁡x−12(x22−x)−12log⁡(x+1)+14x2+C=\frac{x^{2}}{2} \log (x+1)-\frac{x^{2}}{2} \log x-\frac{1}{2}\left(\frac{x^{2}}{2}-x\right)-\frac{1}{2} \log (x+1)+\frac{1}{4} x^{2}+C
=x22log⁡(x+1)−x22log⁡x−12log⁡(x+1)+12x+C=\frac{x^{2}}{2} \log (x+1)-\frac{x^{2}}{2} \log x-\frac{1}{2} \log (x+1)+\frac{1}{2} x+C
Hence, f(x)=x22−12,g(x)=−12log⁡xf(x)=\frac{x^{2}}{2}-\frac{1}{2}, g(x)=-\frac{1}{2} \log x and A=12A=\frac{1}{2}