Question

Find the value of the following integral - $\int {\left( {\dfrac{{\log x}}{{{{(x + 1)}^2}}}dx} \right)}$

Answer 1

In this question we need to find out the integration of $\int {\left( {\dfrac{{\log x}}{{{{(x + 1)}^2}}}dx} \right)}$, for that we need to use integration by parts whose formula is $\int {uvdx = u\int {vdx - \int {\left[ {\dfrac{{du}}{{dx}}\left( {\int {vdx} } \right)} \right]} dx} }$. Using the ILATE rule, we will choose $u = \log x$ and $v = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$. After this we will use the partial fraction method to find out the complete integration.

Complete step-by-step answer:
We have been provided to find the integration of: $\int {\left( {\dfrac{{\log x}}{{{{(x + 1)}^2}}}dx} \right)}$,
For that we will let log x equal to the first term and $\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$ as second term,
Now we will be using integration by parts method in order to find the integration of $\int {\left( {\dfrac{{\log x}}{{{{(x + 1)}^2}}}dx} \right)}$ whose formula is: $\int {uvdx = u\int {vdx - \int {\left[ {\dfrac{{du}}{{dx}}\left( {\int {vdx} } \right)} \right]} dx} }$,
Here, we will make use of the ILATE rule for choosing u and v. The order of choosing the first function is to be considered as Inverse, Logarithm, Algebra, Trigonometry and Exponent, which is the expansion of ILATE. So, here we have logarithmic and algebraic functions. So, as per rule, we can say that $u = \log x$ and $v = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$.
Now keeping the values in the formula,
$I = \log x\int {\dfrac{1}{{{{(x + 1)}^2}}}} dx - \int {\left( {\dfrac{1}{x}.\int {\dfrac{1}{{{{(x + 1)}^2}}}dx} } \right)} dx$, here I represent the integration of $\int {\left( {\dfrac{{\log x}}{{{{(x + 1)}^2}}}dx} \right)}$,
Here, we need to find the integration of $\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$ ,
So, $\int {\dfrac{1}{{{{(x + 1)}^2}}}dx = \dfrac{{ - 1}}{{(x + 1)}}}$ using the formula: $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$,
Now keeping the value of integration of $\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$in $I = \log x\int {\dfrac{1}{{{{(x + 1)}^2}}}} dx - \int {\left( {\dfrac{1}{x}.\int {\dfrac{1}{{{{(x + 1)}^2}}}dx} } \right)} dx$,
So, $I = \log x\left( {\dfrac{{ - 1}}{{(x + 1)}}} \right) + \int {\left( {\dfrac{1}{x}.\left( {\dfrac{1}{{(x + 1)}}} \right)} \right)dx}$,
Now simplifying it further,
$I = \left( {\dfrac{{ - \log x}}{{(x + 1)}}} \right) + \int {\left( {\dfrac{1}{x}.\left( {\dfrac{1}{{(x + 1)}}} \right)} \right)dx}$,
Now we will let $\int {\left( {\dfrac{1}{x}.\left( {\dfrac{1}{{(x + 1)}}} \right)} \right)dx}$ as ${I_1}$, as we need to integrate it again,
So, now the equation becomes: $I = \left( {\dfrac{{ - \log x}}{{(x + 1)}}} \right) + {I_1}$,
Now we need to find ${I_1}$,
So, ${I_1} = \int {\left( {\dfrac{1}{x}.\left( {\dfrac{1}{{(x + 1)}}} \right)} \right)} dx$,
We will be finding the integration of $\int {\left( {\dfrac{1}{x}.\left( {\dfrac{1}{{(x + 1)}}} \right)} \right)} dx$, using partial fraction method,
The formula of partial fraction that we will be using is:$\dfrac{{px + q}}{{(x - a)(x - b)}} = \dfrac{A}{{x - a}} + \dfrac{B}{{x - b}}$where $a \ne b$,
So, using the formula of partial fraction, $\dfrac{1}{{x(x + 1)}} = \dfrac{A}{x} + \dfrac{B}{{x + 1}}$,
Now multiplying both sides by x(x+1),
So, the value comes out to be: $1 = A(x + 1) + B(x)$,
Now on simplifying,$1 = (A + B)x + A$,
Now on comparing the terms,$A + B = 0$ and $A = 1$,
Keeping the value of A in the above equation,$1 + B = 0$,
The value of B comes out to be:$B = - 1$,
Now we fill be finding the value of ${I_1}$, keeping the values of A and B in the formula: :$\dfrac{{px + q}}{{(x - a)(x - b)}} = \dfrac{A}{{x - a}} + \dfrac{B}{{x - b}}$,
So, ${I_1} = \int {\left( {\dfrac{1}{x} - \dfrac{1}{{(x + 1)}}} \right)} dx$,
Now solving it further,${I_1} = \int {\dfrac{1}{x}dx} - \int {\dfrac{1}{{(x + 1)}}dx}$
So, for the integration we will be using the formula:$\int {\dfrac{1}{x}dx = \log x}$,
So, ${I_1} = \log x - \log (x + 1)$,
Now keeping the value of ${I_1}$ in:$I = \left( {\dfrac{{ - \log x}}{{(x + 1)}}} \right) + {I_1}$,
So, the value of I would be:$I = \dfrac{{ - \log x}}{{(x + 1)}} + \log x - \log (x + 1)$.

Note: In this question while doing the integration of $\dfrac{1}{x}$, do not use the formula: $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$,
As it will result in 0 in the denominator which is equal to infinity. So, use the formula: $\int {\dfrac{1}{x}dx = \log x}$to avoid this.