What is the solution for (1+2t)/(1+2t^2)dt= 1/u du ?

The solution for (1+2t)/(1+2t^2)dt= 1/u du is u=e^{1/(sqrt(2))arctan(sqrt(2)t)+c_{1}}sqrt(2t^2+1)

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(1+2t)/(1+2t^2)dt= 1/u du

\frac{1 + 2 t}{1 + 2 t ^{2}} d t = \frac{1}{u} d u

u = e^{\frac{1}{2} a r c t a n (2 t) + c_{1}} 2 t^{2} + 1

Solution steps

\frac{1 + 2 t}{1 + 2 t ^{2}} d t = \frac{1}{u} d u

Solve separable ODE:

u = e^{\frac{1}{2} a r c t a n (2 t) + c_{1}} 2 t^{2} + 1

u = e^{\frac{1}{2} a r c t a n (2 t) + c_{1}} 2 t^{2} + 1

f^{^{'}} (t) - f (t) = 2, f (0) = 0

y y^{^{'}} = x - x y

\frac{d y}{d t} = \frac{e ^{t}}{2 y}

\frac{d y}{d x} = 4 y^{2}, ln (x) y (1) = - 1

y \frac{d y}{d x} + \frac{1}{1 - x ^{2}} (y^{2} + 1) = 0

What is the solution for (1+2t)/(1+2t^2)dt= 1/u du ?
The solution for (1+2t)/(1+2t^2)dt= 1/u du is u=e^{1/(sqrt(2))arctan(sqrt(2)t)+c_{1}}sqrt(2t^2+1)