# How to solve:- evaluate int(dx)/((x+1)(2x+1))

Welcome to my article evaluate int(dx)/((x+1)(2x+1)). This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
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Let us know how to solve this question evaluate int(dx)/((x+1)(2x+1)) .

First write the question on the page of the notebook

## evaluate int(dx)/((x+1)(2x+1))

By writing this question correctly in this way

\displaystyle \operatorname{int}.\frac{{dx}}{{\left( {x+1} \right)\left( {2x+1} \right)}}

\displaystyle \int{{\frac{{dx}}{{\left( {x+1} \right)\left( {2x+1} \right)}}}}

\displaystyle I=\int{{\frac{{dx}}{{\left( {x+1} \right)\left( {2x+1} \right)}}}}

\displaystyle I=\int{{\frac{{\frac{1}{2}}}{{\left( {x+1} \right)\left( {2x+1} \right)}}}}dx

\displaystyle \frac{1}{2}=\left( {x+1} \right)-\left( {x+\frac{1}{2}} \right)

\displaystyle \frac{1}{2}=x+1-x-\frac{1}{2}

\displaystyle I=\int{{\frac{{\left( {x+1} \right)-\left( {x+\frac{1}{2}} \right)}}{{\left( {x+1} \right)\left( {x+\frac{1}{2}} \right)}}}}dx

\displaystyle I=\int{{\frac{{\left( {x+1} \right)}}{{\left( {x+1} \right)\left( {x+\frac{1}{2}} \right)}}}}dx-\int{{\frac{{\left( {x+\frac{1}{2}} \right)}}{{\left( {x+1} \right)\left( {x+\frac{1}{2}} \right)}}}}dx

\displaystyle I=\int{{\frac{{dx}}{{\left( {x+\frac{1}{2}} \right)}}}}-\int{{\frac{{dx}}{{\left( {x+1} \right)}}}}

AND

## evaluate int(dx)/((x+1)(2x+1))

By writing this question correctly in this way.

\displaystyle Evaluate.\int{{\frac{{dx}}{{\left( {x+1} \right)\left( {2x+1} \right)}}}}

\displaystyle \int{{\frac{{dx}}{{\left( {x+1} \right)\left( {2x+1} \right)}}}}

\displaystyle \int{{\frac{{2\left( {x+1} \right)-\left( {2x+1} \right)}}{{\left( {x+1} \right)\left( {2x+1} \right)}}dx}}

\displaystyle 2\int{{\frac{{dx}}{{\left( {2x+1} \right)}}-\int{{\frac{{dx}}{{\left( {x+1} \right)}}}}}}

\displaystyle 2\int{{\frac{{d\left( {2x+1} \right)}}{{\left( {2x+1} \right)}}-\int{{\frac{{d\left( {x+1} \right)}}{{\left( {x+1} \right)}}}}}}

FORMULA :-

\displaystyle {{\int{{\frac{{dx}}{x}=\log }}}_{e}}x+c

\displaystyle \log e\left| {2x+1} \right|-\log e\left| {x+1} \right|+c

Note :- [c = Constant of the Integration ]

## \displaystyle {{\log }_{e}}\left| {\frac{{2x+1}}{{x+1}}} \right|+c [Answer]

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