Question

integral
2x +7/2x²7x+3 dx​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{2x+7}{2x^2+7x+3}\ dx$

Multiplying and dividing by 2,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{4x+14}{2x^2+7x+3}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{4x+7+7}{2x^2+7x+3}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{4x+7}{2x^2+7x+3}\ dx+\dfrac{7}{2}\int\dfrac{1}{2x^2+7x+3}\ dx\quad\quad\dots(1)$

Consider,

$\displaystyle\longrightarrow I_1=\int\dfrac{4x+7}{2x^2+7x+3}\ dx$

$\displaystyle\longrightarrow I_1=\log\left|2x^2+7x+3\right|$

Consider,

$\displaystyle\longrightarrow I_2=\int\dfrac{1}{2x^2+7x+3}\ dx$

$\displaystyle\longrightarrow I_2=\int\dfrac{1}{2x^2+6x+x+3}\ dx$

$\displaystyle\longrightarrow I_2=\int\dfrac{1}{2x(x+3)+(x+3)}\ dx$

$\displaystyle\longrightarrow I_2=\int\dfrac{1}{(x+3)(2x+1)}\ dx$

$\displaystyle\longrightarrow I_2=2\int\dfrac{1}{(2x+6)(2x+1)}\ dx$

$\displaystyle\longrightarrow I_2=\dfrac{2}{5}\int\dfrac{(2x+6)-(2x+1)}{(2x+6)(2x+1)}\ dx$

$\displaystyle\longrightarrow I_2=\dfrac{2}{5}\int\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+6}\right)\ dx$

$\displaystyle\longrightarrow I_2=\dfrac{1}{5}\int\dfrac{2\ dx}{2x+1}-\dfrac{1}{5}\int\dfrac{dx}{x+3}$

$\displaystyle\longrightarrow I_2=\dfrac{1}{5}\log\left|\dfrac{2x+1}{x+3}\right|$

Then (1) becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\log\left|2x^2+7x+3\right|+\dfrac{7}{2}\cdot\dfrac{1}{5}\log\left|\dfrac{2x+1}{x+3}\right|+C$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\log\left|2x^2+7x+3\right|+\dfrac{7}{10}\log\left|\dfrac{2x+1}{x+3}\right|+C$

$\displaystyle\longrightarrow I=\dfrac{5}{10}\log\left|(2x+1)(x+3)\right|+\dfrac{7}{10}\log\left|\dfrac{2x+1}{x+3}\right|+C$

$\displaystyle\longrightarrow I=\dfrac{5}{10}\log\left|2x+1\right|+\dfrac{5}{10}\log\left|x+3\right|+\dfrac{7}{10}\log|2x+1|-\dfrac{7}{10}\log|x+3|+C$

$\displaystyle\longrightarrow I=\dfrac{6}{5}\log\left|2x+1\right|-\dfrac{1}{5}\log|x+3|+C$

$\displaystyle\longrightarrow\underline{\underline{I=\log\left|\dfrac{(2x+1)^6}{x+3}\right|^{\frac{1}{5}}+C}}$