Answer
Verified
91.5k+ views
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If the distance between 1st crest and the third crest class 11 physics JEE_Main
A man of mass 50kg is standing on a 100kg plank kept class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main