What is the solution for y^'=arctan(x/2) ?

The solution for y^'=arctan(x/2) is y=xarctan(x/2)-ln(x^2+4)+c_{1}

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y^'=arctan(x/2)

y^{^{'}} = arctan (\frac{x}{2})

y = x arctan (\frac{x}{2}) - ln (x^{2} + 4) + c_{1}

Solution steps

y^{'} (x) = arctan (\frac{x}{2})

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int arctan (\frac{x}{2}) d x

\int arctan (\frac{x}{2}) d x = x arctan (\frac{x}{2}) - ln (x^{2} + 4) + c_{1}

y = x arctan (\frac{x}{2}) - ln (x^{2} + 4) + c_{1}

\int x \cdot e^{- 5 x} d x

\frac{d}{d x} (5^{5 x})

t \to 0 lim (e^{- 7 t} i + \frac{t ^{2}}{sin ^{2} ( t )} j + tan (3 t) k)

\int \frac{1}{3} tan (3 x) d x

\int cos^{2} (θ) - sin^{2} (θ) d θ

What is the solution for y^'=arctan(x/2) ?
The solution for y^'=arctan(x/2) is y=xarctan(x/2)-ln(x^2+4)+c_{1}