If = $\int x \log\left(1+ \frac{1}{x}\right)dx = f\left(x\right)\log\left(x+1\right)+g\left(x\right)x^{2}+Lx +C$, then
- $f(x) = \frac{1}{2} x^2$
- $g(x) = \log x$
- $L = 1$
- None of these
The Correct Option is D
Solution and Explanation
$=\int x \log (x+1) d x-\int x \log x d x$
$=\frac{x^{2}}{2} \log (x+1)-\frac{1}{2} \int \frac{x^{2}}{x+1} d x-\frac{x^{2}}{2} \log x+\frac{1}{2} \int \frac{x^{2}}{x} d x$
$=\frac{x^{2}}{2} \log (x+1)-\frac{1}{2} \int\left(x-1+\frac{1}{x+1}\right) d x-\frac{x^{2}}{2} \log x+\frac{1}{4} x^{2}$
$=\frac{x^{2}}{2} \log (x+1)-\frac{x^{2}}{2} \log x-\frac{1}{2}\left(\frac{x^{2}}{2}-x\right)-\frac{1}{2} \log (x+1)+\frac{1}{4} x^{2}+C$
$=\frac{x^{2}}{2} \log (x+1)-\frac{x^{2}}{2} \log x-\frac{1}{2} \log (x+1)+\frac{1}{2} x+C$
Hence, $f(x)=\frac{x^{2}}{2}-\frac{1}{2}, g(x)=-\frac{1}{2} \log x$ and $A=\frac{1}{2}$
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C
Integration Using Trigonometric Identities
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