Question

Find $\int x^2+a^2$ and evaluate $\int \frac{1}{3+2x+x^2}dx$

Answer 1

$f(x^2+a^2)dx$

$=\frac{x^3}{3}+a^{2}x+c$

$\therefore f(x^2+a^2)dx=\frac{x^3}{3}+a^{2}x+c$

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

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

$=\int \frac{dx}{{(x+1)}^2+{\left(\sqrt{2}\right)}^2}$

$=\frac{1}{\sqrt{2}}\tan -1\left(\frac{x+1}{\sqrt{2}}\right)+c\left[\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan -1\left(\frac{x}{a}\right)\right]$

$\therefore \int \frac{dx}{x^2+2x+3}=\frac{1}{\sqrt{2}}\tan -1\left(\frac{x+1}{\sqrt{2}}\right)+c$