Question

integration 2x+7/
2x²+x+3 dx​

Answer 1

We're asked to evaluate the integral,

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

Multiplying and dividing by 2,

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

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{4x+1+13}{2x^2+x+3}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{4x+1}{2x^2+x+3}\ dx+\dfrac{13}{2}\int\dfrac{1}{2x^2+x+3}\ dx\quad\quad\dots(1)$

Consider,

$\displaystyle\longrightarrow I_1=\int\dfrac{4x+1}{2x^2+x+3}\ dx$

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

Consider,

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

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

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\int\dfrac{1}{\left(x^2+\dfrac{1}{2}\,x+\dfrac{1}{16}+\dfrac{23}{16}\right)}\ dx$

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\int\dfrac{1}{\left(x+\dfrac{1}{4}\right)^2+\left(\dfrac{\sqrt{23}}{4}\right)^2}\ dx$

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\cdot\dfrac{1}{\left(\frac{\sqrt{23}}{4}\right)}\tan^{-1}\left(\dfrac{x+\frac{1}{4}}{\frac{\sqrt{23}}{4}}\right)$

$\displaystyle\longrightarrow I_2=\dfrac{2}{\sqrt{23}}\tan^{-1}\left(\dfrac{4x+1}{\sqrt{23}}\right)$

Then (1) becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\log\left|2x^2+x+3\right|+\dfrac{13}{2}\cdot\dfrac{2}{\sqrt{23}}\tan^{-1}\left(\dfrac{4x+1}{\sqrt{23}}\right)+C$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}\log\left|2x^2+x+3\right|+\dfrac{13}{\sqrt{23}}\tan^{-1}\left(\dfrac{4x+1}{\sqrt{23}}\right)+C}}$