What is the solution for (dr)/(dt)= 1/(t+1),r(0)=1 ?

The solution for (dr)/(dt)= 1/(t+1),r(0)=1 is r=ln(t+1)+1

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(dr)/(dt)= 1/(t+1),r(0)=1

\frac{d r}{d t} = \frac{1}{t + 1}, r (0) = 1

r = ln (t + 1) + 1

Solution steps

\frac{d r}{d t} = \frac{1}{t + 1}

Substitute

\frac{d r}{d t}

with

r^{'} (t)

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

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

then

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

r = \int \frac{1}{t + 1} d t

\int \frac{1}{t + 1} d t = ln (t + 1) + c_{1}

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

Apply initial conditions:

r = ln (t + 1) + 1

r = ln (t + 1) + 1

(1 + x^{2}) y^{^{'}} + 1 + 2 x y = 0

\int x^{2} (3 - e^{x^{3}}) d x

x \to \infty lim (e^{- \frac{2}{x}})

\frac{2}{s}

\int_{1}^{2} 0 d x

What is the solution for (dr)/(dt)= 1/(t+1),r(0)=1 ?
The solution for (dr)/(dt)= 1/(t+1),r(0)=1 is r=ln(t+1)+1