[tex]f'(x)=\dfrac{4x}{1+7x^2}[/tex]
Integrating gives
[tex]f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx[/tex]
To compute the integral, substitute [tex]u=1+7x^2[/tex], so that [tex]\frac27\,\mathrm du=4x\,\mathrm dx[/tex]. Then
[tex]f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C[/tex]
Since [tex]u=1+7x^2>0[/tex] for all [tex]x[/tex], we can drop the absolute value, so we end up with
[tex]f(x)=\dfrac27\ln(1+7x^2)+C[/tex]
Given that [tex]f(0)=10[/tex], we have
[tex]10=\dfrac27\ln1+C\implies C=10[/tex]
so that
[tex]\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}[/tex]
