Answer
$$\int_{0}^{1}\frac{2}{2x^{2}+3x+1}dx=ln\frac{9}{4}$$
Work Step by Step
$$\frac{2}{2x^{2}+3x+1}=\frac{2}{(2x+1)(x+1)}$$
$$=\frac{A}{2x+1}+\frac{B}{x+1}=\frac{(A+B)+(A+2B)x}{(2x+1)(x+1)}$$
$$A=4,B=-2$$
$$\int_{0}^{1}\frac{2}{2x^{2}+3x+1}dx=\int_{0}^{1}(\frac{4}{2x+1}-\frac{2}{x+1})dx$$
$$=\left [ 2ln|2x+1|-2ln|x+1| \right ]_{0}^{1}$$
$$=2ln3-2ln2=ln\frac{9}{4}$$
