Evaluate
x\sqrt{y^{2}+16}+С
Differentiate w.r.t. x
\sqrt{y^{2}+16}
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\sqrt{16+y^{2}}x
Find the integral of \sqrt{16+y^{2}} using the table of common integrals rule \int a\mathrm{d}x=ax.
\sqrt{16+y^{2}}x+С
If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.
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