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$\int {\dfrac{{1 + \log x}}{{x\left( {2 + \log x} \right)\left( {3 + \log x} \right)}}} {\text{dx}}$

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Hint- First use substitution method and then partial fraction method to simplify the integral.

Here we have to evaluate \[\int {\dfrac{{1 + \log x}}{{x\left( {2 + \log x} \right)\left( {3 + \log x} \right)}}} {\text{dx}}\]

So let’s substitute \[{\text{1 + logx = p}}\]

So on differentiating both the sides we have \[\left( {{\text{0 + }}\dfrac{1}{x}} \right)dx = dp\]

Let’s make this substitution back into our main integral we get

\[\int {\dfrac{p}{{x\left( {1 + p} \right)\left( {2 + p} \right)}} \times xdp} \] As \[{\text{(2 + logx)}}\]can be written as \[{\text{(1 + (1 + logx))}}\]and \[\left( {3 + \log x} \right)\]can be written as \[\left( {2 + (1 + \log x)} \right)\]

On simplifying we get

\[\int {\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}}} {\text{dp}}\]

Now let’s resolve it into partial fractions so we can write this form as

Let \[\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{A}{{1 + p}} + \dfrac{B}{{2 + p}}\]……………………………. (1)

So let’s take LCM in the right side we get

\[\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{{A\left( {2 + p} \right) + B\left( {1 + p} \right)}}{{\left( {1 + p} \right)\left( {2 + p} \right)}}\]

Denominator in both sides will cancel it out so we get

\[{\text{p = }}A\left( {2 + p} \right) + B\left( {1 + p} \right)\]……………………………………. (2)

Now let’s put \[{\text{p = - 2}}\] so that we can find the value of B

We get

\[{\text{ - 2 = 0 - B}}\] Hence our \[{\text{B = 2}}\]………………………………………. (3)

Now we need to find A so let’s put \[{\text{p = - 1}}\] in equation (2) we get

\[{\text{ - 1 = A + 0}}\] Hence our \[{\text{A = - 1}}\]………………………………… (4)

Now let’s put equation (3) and equation (4) back into (1) we get

\[\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{{ - 1}}{{1 + p}} + \dfrac{2}{{2 + p}}\]

So our integral can now be written as

\[{\text{I}} = \int {\left( {\dfrac{{ - 1}}{{1 + p}} + \dfrac{2}{{2 + p}}} \right){\text{ }}} {\text{dp}}\]

We can segregate this integral as

\[{\text{I = }}\int {\dfrac{{ - 1}}{{1 + p}}{\text{ dp}}} {\text{ + }}\int {\dfrac{2}{{2 + p}}} {\text{ dp}}\]

Now using the standard formula for integral of\[\int {\dfrac{{dx}}{x} = \log \left| x \right|} \], we can solve above as

\[{\text{I = - log}}\left| {1 + p} \right|{\text{ + 2log}}\left| {2 + p} \right|{\text{ + c}}\]

Let’s substitute back the value of p which was \[{\text{1 + logx}}\] we get

\[{\text{I = - log}}\left| {2 + \log x} \right| + 2\log \left| {3 + \log x} \right|{\text{ + c}}\]

Using the property of log that is \[{\text{log(A) - log(B) = log}}\left( {\dfrac{A}{B}} \right)\]and \[{\text{nlog(x) = log(x}}{{\text{)}}^n}\]

\[{\text{I = log}}\left| {\dfrac{{{{\left( {\log x + 3} \right)}^2}}}{{\log x + 2}}} \right|{\text{ + c}}\]

Note- All such type of problems are based upon the concept of evaluation by substitution, whenever we see some terms in the integral that are related or can be converted into one another in numerator or in denominator then we can use the basic term and substitute it to some quantity in order to simplify the integral.

Here we have to evaluate \[\int {\dfrac{{1 + \log x}}{{x\left( {2 + \log x} \right)\left( {3 + \log x} \right)}}} {\text{dx}}\]

So let’s substitute \[{\text{1 + logx = p}}\]

So on differentiating both the sides we have \[\left( {{\text{0 + }}\dfrac{1}{x}} \right)dx = dp\]

Let’s make this substitution back into our main integral we get

\[\int {\dfrac{p}{{x\left( {1 + p} \right)\left( {2 + p} \right)}} \times xdp} \] As \[{\text{(2 + logx)}}\]can be written as \[{\text{(1 + (1 + logx))}}\]and \[\left( {3 + \log x} \right)\]can be written as \[\left( {2 + (1 + \log x)} \right)\]

On simplifying we get

\[\int {\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}}} {\text{dp}}\]

Now let’s resolve it into partial fractions so we can write this form as

Let \[\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{A}{{1 + p}} + \dfrac{B}{{2 + p}}\]……………………………. (1)

So let’s take LCM in the right side we get

\[\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{{A\left( {2 + p} \right) + B\left( {1 + p} \right)}}{{\left( {1 + p} \right)\left( {2 + p} \right)}}\]

Denominator in both sides will cancel it out so we get

\[{\text{p = }}A\left( {2 + p} \right) + B\left( {1 + p} \right)\]……………………………………. (2)

Now let’s put \[{\text{p = - 2}}\] so that we can find the value of B

We get

\[{\text{ - 2 = 0 - B}}\] Hence our \[{\text{B = 2}}\]………………………………………. (3)

Now we need to find A so let’s put \[{\text{p = - 1}}\] in equation (2) we get

\[{\text{ - 1 = A + 0}}\] Hence our \[{\text{A = - 1}}\]………………………………… (4)

Now let’s put equation (3) and equation (4) back into (1) we get

\[\dfrac{p}{{\left( {1 + p} \right)\left( {2 + p} \right)}} = \dfrac{{ - 1}}{{1 + p}} + \dfrac{2}{{2 + p}}\]

So our integral can now be written as

\[{\text{I}} = \int {\left( {\dfrac{{ - 1}}{{1 + p}} + \dfrac{2}{{2 + p}}} \right){\text{ }}} {\text{dp}}\]

We can segregate this integral as

\[{\text{I = }}\int {\dfrac{{ - 1}}{{1 + p}}{\text{ dp}}} {\text{ + }}\int {\dfrac{2}{{2 + p}}} {\text{ dp}}\]

Now using the standard formula for integral of\[\int {\dfrac{{dx}}{x} = \log \left| x \right|} \], we can solve above as

\[{\text{I = - log}}\left| {1 + p} \right|{\text{ + 2log}}\left| {2 + p} \right|{\text{ + c}}\]

Let’s substitute back the value of p which was \[{\text{1 + logx}}\] we get

\[{\text{I = - log}}\left| {2 + \log x} \right| + 2\log \left| {3 + \log x} \right|{\text{ + c}}\]

Using the property of log that is \[{\text{log(A) - log(B) = log}}\left( {\dfrac{A}{B}} \right)\]and \[{\text{nlog(x) = log(x}}{{\text{)}}^n}\]

\[{\text{I = log}}\left| {\dfrac{{{{\left( {\log x + 3} \right)}^2}}}{{\log x + 2}}} \right|{\text{ + c}}\]

Note- All such type of problems are based upon the concept of evaluation by substitution, whenever we see some terms in the integral that are related or can be converted into one another in numerator or in denominator then we can use the basic term and substitute it to some quantity in order to simplify the integral.