Question

Integrate $\int {\dfrac{{\log x}}{{x\left( {1 + \log x} \right)\left( {2 + \log x} \right)}}} dx$.

Answer 1

To solve this question, we will assume it to be equal to $I$. Then we will make a substitution in the given question and then change the integrand accordingly. After substitution we will get an integrand that we will solve by partial fraction decomposition method.

Complete step by step answer:
Let,
$I = \int {\dfrac{{\log x}}{{x\left( {1 + \log x} \right)\left( {2 + \log x} \right)}}} dx$
Now we can see that there is $\log x$ in the numerator and $x$ in the denominator. Also, we know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$. So, we will make a substitution for $\log x$.
Let, $u = \log x$
Now we will differentiate both sides. So, we get;
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\log x$
Solving we get;
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{x}$
From here we will find $dx$, and replace $dx$ in the main integration. So, we get;
$\Rightarrow dx = xdu$
Now using these we will write $I$ again. So, we get;
$I = \int {\dfrac{u}{{x\left( {1 + u} \right)\left( {2 + u} \right)}}} xdu$
Cancelling $x$ from the numerator and denominator we get;
$\Rightarrow I = \int {\dfrac{u}{{\left( {1 + u} \right)\left( {2 + u} \right)}}} du$
Now we will decompose $\dfrac{u}{{\left( {1 + u} \right)\left( {2 + u} \right)}}$ into a partial fraction. So, we get;
$\Rightarrow \dfrac{u}{{\left( {1 + u} \right)\left( {2 + u} \right)}} = \dfrac{A}{{u + 1}} + \dfrac{B}{{u + 2}}$
Now we will take LCM and solve it further. So, we get;
$\Rightarrow \dfrac{u}{{\left( {1 + u} \right)\left( {2 + u} \right)}} = \dfrac{{A\left( {u + 2} \right) + B\left( {u + 1} \right)}}{{\left( {1 + u} \right)\left( {2 + u} \right)}}$
Now we will cancel the denominators of both sides. So, we get;
$\Rightarrow u = A\left( {u + 2} \right) + B\left( {u + 1} \right)$
Now we can see from the denominator of the above equation when we find the roots by equating them to zero, we get;
$u + 2 = 0$
$\Rightarrow u = - 2$
Also,
$u + 1 = 0$
$\Rightarrow u = - 1$
Now we will put these values in the above equation one by one to get the value of constant. So, putting $u = - 2$, we have;
$\Rightarrow - 2 = A\left( { - 2 + 2} \right) + B\left( { - 2 + 1} \right)$
We can see that the first term on the RHS is zero. So, we get;
$\Rightarrow - 2 = B\left( { - 1} \right)$
On solving we get;
$\Rightarrow B = 2$
Now we will put $u = - 1$, so, we get;
$\Rightarrow - 1 = A\left( { - 1 + 2} \right) + B\left( { - 1 + 1} \right)$
Solving we get;
$\Rightarrow - 1 = A\left( 1 \right)$
Solving we get;
$\Rightarrow A = - 1$
So, putting the value of constant we get;
$\Rightarrow \dfrac{u}{{\left( {u + 1} \right)\left( {u + 2} \right)}} = \dfrac{{ - 1}}{{\left( {u + 1} \right)}} + \dfrac{2}{{\left( {u + 2} \right)}}$
Now we will put this value in the main integration so, we get;
$\Rightarrow I = \int {\left( {\dfrac{{ - 1}}{{\left( {u + 1} \right)}} + \dfrac{2}{{\left( {u + 2} \right)}}} \right)du}$
Now we will separate the terms, so, we get;
$\Rightarrow I = \int {\dfrac{{ - 1}}{{\left( {u + 1} \right)}}} du + \int {\dfrac{2}{{\left( {u + 2} \right)}}du}$
Now we know $\int {\dfrac{1}{{\left( {u + 1} \right)}}} du = \log |u + 1|$, so, we get;
$\Rightarrow I = - \log |u + 1| + 2\log |u + 2| + C$
Now we will put the value of $u = \log x$. So, we get;
$\Rightarrow I = - \log |\log x + 1| + 2\log |\log x + 2| + C$

Note: One major mistake that students make is that they forget to put the constant in the final answer. As this is the case of indefinite integral, we will have to put the constant otherwise our answer will be wrong. Also, as this is an indefinite integral so, after substitution there is no problem of changing the limits. But in any definite integral question, after substitution we need to change the limits accordingly.