How would I solve this?

[tex]A=\frac{1}{2}\ln17 = 1.417[/tex]Step-by-step explanation:The area A under the curve can be written as[tex]\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2}[/tex]To evaluate the integral, let[tex]u = 1+4x^2 \Rightarrow du = 8xdx\:\text{or}\:\frac{1}{2}du = 4xdx[/tex]so the integral becomes[tex]\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\int\!\dfrac{du}{u} = \dfrac{1}{2}\ln |u|[/tex]or[tex]\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\ln |1+4x^2|[/tex]Putting in the limits of integration, our area becomes [tex]\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\left.\ln |1+4x^2|\right|_0^2[/tex][tex]\;\;\;\;= \frac{1}{2}[\ln (1+16) - \ln (1)][/tex][tex]\;\;\;\;=\frac{1}{2}\ln17[/tex]Note: [tex]\ln 1 = 0[/tex]