Question

integral x^2+3x÷x^2+1

Answer 1

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

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

$=\displaystyle x+\dfrac{3}{2}\int\dfrac{2x}{x^2+1}\,dx-\dfrac{3}{2}\int\dfrac{\frac{2}{3}}{x^2+1}\,dx\\\\\\ =x+\dfrac{3}{2}\int\dfrac{2x}{x^2+1}\,dx-\int\dfrac{1}{x^2+1}\,dx\\\\\\ =x+\dfrac{3}{2}\,\mathrm{\ell n}(x^2+1)-\mathrm{arctg\,}x+C$

$\therefore~~\boxed{\begin{array}{c} \displaystyle\int \dfrac{x^2+3x}{x^2+1}\,dx=x+\dfrac{3}{2}\,\mathrm{\ell n}(x^2+1)-\mathrm{arctg\,}x+C \end{array}}$