What is the solution for (dr)/(dt)=(sec^2(t))/(tan(t)+1) ?

The solution for (dr)/(dt)=(sec^2(t))/(tan(t)+1) is r=ln(tan(t)+1)+c_{1}

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(dr)/(dt)=(sec^2(t))/(tan(t)+1)

\frac{d r}{d t} = \frac{sec ^{2} ( t )}{tan ( t ) + 1}

r = ln (tan (t) + 1) + c_{1}

Solution steps

\frac{d r}{d t} = \frac{sec ^{2} ( t )}{tan ( t ) + 1}

Substitute

\frac{d r}{d t}

with

r^{'} (t)

r^{^{'}} (t) = \frac{sec ^{2} ( t )}{tan ( t ) + 1}

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

then

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

r = \int \frac{sec ^{2} ( t )}{tan ( t ) + 1} d t

\int \frac{sec ^{2} ( t )}{tan ( t ) + 1} d t = ln (tan (t) + 1) + c_{1}

r = ln (tan (t) + 1) + c_{1}

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What is the solution for (dr)/(dt)=(sec^2(t))/(tan(t)+1) ?
The solution for (dr)/(dt)=(sec^2(t))/(tan(t)+1) is r=ln(tan(t)+1)+c_{1}