Question

Solve the integration \[\int {\dfrac{{3x + 1}}{{2{x^2} + x + 1}}} dx = ................\]
A) \[\dfrac{3}{4}\log (2{x^2} + x + 1) + \dfrac{1}{{2\sqrt {\left( 7 \right)} }}{\tan ^{ - 1}}\dfrac{{4x + 1}}{{\sqrt {\left( 7 \right)} }} + C\].
B) \[\dfrac{4}{3}\log (2{x^2} + x + 1) + \dfrac{1}{{2\sqrt {\left( 7 \right)} }}{\tan ^{ - 1}}\dfrac{{3x + 1}}{{\sqrt {\left( 7 \right)} }} + C\].
C) \[\dfrac{4}{3}\log (2{x^2} + x + 1) + \dfrac{1}{{2\sqrt {\left( 7 \right)} }}{\cot ^{ - 1}}\dfrac{{3x + 1}}{{\sqrt {\left( 7 \right)} }} + C\]
D) None of these

Answer 1

Hint: First we have to know the integration is the inverse process of the differentiation. The given integrand can be integrated by substitution method. First we have to convert the integrand in the form such that it can be integral by using any integration methods.

Complete step by step solution:
Given \[\int {\dfrac{{3x + 1}}{{2{x^2} + x + 1}}} dx\]
Let\[I = \int {\dfrac{{3x + 1}}{{2{x^2} + x + 1}}} dx\] --(1)
Multiply and divided by \[3\] in the integrand of the equation (1), we get
\[I = \int {\left( {\dfrac{3}{3} \times \dfrac{{3x + 1}}{{2{x^2} + x + 1}}} \right)} dx\]
\[ \Rightarrow I = \int {\left( {3 \times \dfrac{{\left( {x + \dfrac{1}{3}} \right)}}{{2{x^2} + x + 1}}} \right)} dx\]--(2)
Again, multiply and divided by \[4\] in the integrand of the equation (2), we get
\[I = \int {\left( {\dfrac{3}{4} \times \dfrac{{\left( {4x + \dfrac{4}{3}} \right)}}{{2{x^2} + x + 1}}} \right)} dx\]----(3)
The equation (3) can be rewritten as
\[I = \int {\left( {\dfrac{3}{4} \times \dfrac{{\left( {4x + 1 - 1 + \dfrac{4}{3}} \right)}}{{2{x^2} + x + 1}}} \right)} dx\]----(4)
Slitting the integrand in the equation (4), we get
\[I = \dfrac{3}{4}\left[ {\int {\left( {\dfrac{{4x + 1}}{{2{x^2} + x + 1}} + \dfrac{1}{{3\left( {2{x^2} + x + 1} \right)}}} \right)} dx} \right]\]----(5)
Since we know that integration of the sum of two functions is equal to the sum of the integration of two functions. Then the equation (5) becomes
\[I = \dfrac{3}{4}\int {\dfrac{{4x + 1}}{{2{x^2} + x + 1}}dx} + \dfrac{1}{4}\int {\left( {\dfrac{1}{{\left( {2{x^2} + x + 1} \right)}}} \right)} dx\].
Let \[I = {I_1} + {I_2}\]-----(6)
Where, \[{I_1} = \dfrac{3}{4}\int {\dfrac{{4x + 1}}{{2{x^2} + x + 1}}dx} \]---(7) and \[{I_2} = \dfrac{1}{4}\int {\left( {\dfrac{1}{{\left( {2{x^2} + x + 1} \right)}}} \right)} dx\]---(8)
Put \[2{x^2} + x + 1 = t\] in the equation (7) and we get \[\left( {4x + 1} \right)dx = dt\]. Then the equation (7) becomes
\[{I_1} = \dfrac{3}{4}\int {\dfrac{1}{t}dt} \]

\[ \Rightarrow \]\[{I_1} = \dfrac{3}{4}\log t + {C_1}\]----(9)
Replace \[t = 2{x^2} + x + 1\] in the equation (9), we get
\[{I_1} = \dfrac{3}{4}\log \left( {2{x^2} + x + 1} \right) + {C_1}\] Where \[{C_1}\] is an integration constant
Multiply and divided by \[2\] in the equation (8), we get
\[{I_2} = \int {\left( {\dfrac{2}{{\left( {16{x^2} + 8x + 8} \right)}}} \right)} dx\]---(10)
Since \[{\left( {4x + 1} \right)^2} = 16{x^2} + 8x + 1\], then the equation (10) becomes
\[{I_2} = \int {\left( {\dfrac{2}{{{{\left( {4x + 1} \right)}^2} + {{\left( {\sqrt 7 } \right)}^2}}}} \right)} dx\]---(11)
Since we know that \[\int {\left( {\dfrac{1}{{{a^2} + {x^2}}}} \right)} dx = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)\], then the equation (11) becomes
\[{I_2} = 2 \times \dfrac{1}{{\sqrt 7 }}{\tan ^{ - 1}}\left( {\dfrac{{4x + 1}}{{\sqrt 7 }}} \right) \times \dfrac{1}{4} + {C_2}\]
\[ \Rightarrow \]\[{I_2} = \dfrac{1}{{2\sqrt 7 }}{\tan ^{ - 1}}\left( {\dfrac{{4x + 1}}{{\sqrt 7 }}} \right) + {C_2}\]------(12)
Where \[{C_2}\] is an integration constant.
Hence the equation (6) becomes
\[I = \dfrac{3}{4}\log \left( {2{x^2} + x + 1} \right) + \dfrac{1}{{2\sqrt 7 }}{\tan ^{ - 1}}\left( {\dfrac{{4x + 1}}{{\sqrt 7 }}} \right) + {C_1} + {C_2}\]
\[ \Rightarrow \]\[\int {\dfrac{{3x + 1}}{{2{x^2} + x + 1}}} dx = \dfrac{3}{4}\log \left( {2{x^2} + x + 1} \right) + \dfrac{1}{{2\sqrt 7 }}{\tan ^{ - 1}}\left( {\dfrac{{4x + 1}}{{\sqrt 7 }}} \right) + C\]
Where \[C = {C_1} + {C_2}\] is an integration constant.
Hence,Option (A) is correct.

Note:
Note that the substitution method used when the derivative of the substitution function must exist in the given integrand. Every integral function is differentiable and continuous but the converse is not true. Similarly, every differentiable function is continuous but the converse is not true.