Question

Evaluate: $\int \frac{3x+1}{2x^2+x+1}dx$

Answer 1

$\int \frac{3x+1}{2x^2+x+1}dx$

$3x+1=A(4x+1)+B$

$3x+1=4Ax+(A+B)$

$4A=3\Rightarrow A=\frac{3}{4}$

$A+B=1$

$B=1-\frac{3}{4}=\frac{1}{4}$

$\int \frac{\frac{3}{4}(4x+1)+\frac{1}{4}}{2x^2+x+1}dx$

$\int \frac{4x+1}{2x^2+x+1}dx+\frac{1}{4}\int \frac{dx}{2x^2+x+1}$

$2x^2+x+1=t$

$(4x+1)dx=dt$

$\frac{3}{4}\int \frac{dt}{f}+\frac{1}{4}\int \frac{dx}{2x^2+x+1}$

$\frac{3}{4}ln|2x^2+x+1|+\frac{1}{4}\int \frac{dx}{2x^2+x+1}$

$\frac{3}{4}ln|2x^2+x+1|+\frac{1}{8}\int \frac{dx}{x^2+\frac{x}{2}+\frac{1}{2}}$

$\frac{3}{4}ln|2x^2+x+1|+\frac{1}{8}\int \frac{dx}{x^2+\frac{x}{2}+\frac{1}{16}+\frac{1}{2}-\frac{1}{16}}$

$\frac{3}{4}ln|2x^2+x+1|+\frac{1}{8}\int \frac{dx}{(x+\frac{1}{4}{)}^2+(\frac{\sqrt{7}}{4}{)}^2}$

$\frac{3}{4}ln|2x^2+x+1|+\frac{1}{8}\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}}\right)+C$

$\frac{3}{4}ln|2x^2+x+1|+\frac{1}{2\sqrt{7}}{\tan }^{-1}\left(\frac{4x+1}{\sqrt{7}}\right)+C$